Ray Optics And Optical Instruments Class 12th Physics Part Ii CBSE Solution

Class 12th Physics Part Ii CBSE Solution
Exercises
  1. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius…
  2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the…
  3. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying…
  4. Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to…
  5. A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What…
  6. A prism is made of glass of unknown refractive index. A parallel beam of light is incident…
  7. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with…
  8. A beam of light converges at a point P. Now a lens is placed in the path of the convergent…
  9. An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm.…
  10. What is the focal length of a convex lens of focal length 30cm in contact with a concave…
  11. A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece…
  12. A person with a normal near point (25cm) using a compound microscope with objective of…
  13. A small telescope has an objective lens of focal length 144cm and an eyepiece of focal…
  14. (a) A giant refracting telescope at an observatory has an objective lens of focal length…
  15. Use the mirror equation to deduce that: (a) an object placed between f and 2f of a concave…
  16. A small pin fixed on a table top is viewed from above from a distance of 50cm. By what…
  17. (a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of…
  18. You have learnt that plane and convex mirrors produce virtual images of objects. Can they…
  19. A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual…
  20. A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would…
  21. Does the apparent depth of a tank of water change if viewed obliquely? If so, does the…
  22. The refractive index of diamond is much greater than that of ordinary glass. Is this fact…
  23. The image of a small electric bulb fixed on the wall of a room is to be obtained on the…
  24. A screen is placed 90cm from an object. The image of the object on the screen is formed by…
  25. Determine the ‘effective focal length’ of the combination of the two lenses in Exercise…
  26. An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a)…
  27. At what angle should a ray of light be incident on the face of a prism of refracting angle…
  28. You are given prisms made of crown glass and flint glass with a wide variety of angles.…
  29. For a normal eye, the far point is at infinity and the near point of distinct vision is…
  30. Does short-sightedness (myopia) or long-sightedness (hyper - metropia) imply necessarily…
  31. A myopic person has been using spectacles of power -1.0 dioptre for distant vision. During…
  32. A person looking at a person wearing a shirt with a pattern comprising vertical and…
  33. A man with normal near point (25 cm) reads a book with small print using a magnifying…
  34. A card sheet divided into squares each of size 1 mm^2 is being viewed at a distance of 9…
  35. At what distance should the lens be held from the figure in Exercise 9.29 in order to view…
  36. What is the magnification in this case?
  37. Is the magnification equal to the magnifying power in this case? Explain.…
  38. What should be the distance between the object in Exercise 9.30 and the magnifying glass…
  39. The angle subtended at the eye by an object is equal to the angle subtended at the eye by…
  40. In viewing through a magnifying glass, one usually positions one’s eyes very close to the…
  41. Magnifying power of a simple microscope is inversely proportional to the focal length of…
  42. Why must both the objective and the eyepiece of a compound microscope have short focal…
  43. When viewing through a compound microscope, our eyes should be positioned not on the…
  44. An angular magnification (magnifying power) of 30X is desired using an objective of focal…
  45. A small telescope has an objective lens of focal length 140cm and an eyepiece of focal…
  46. For the telescope described in Exercise 9.34 (a), what is the separation between the…
  47. If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the…
  48. What is the height of the final image of the tower if it is formed at 25cm?…
  49. A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built…
  50. Light incident normally on a plane mirror attached to a galvanometer coil retraces…
  51. Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid…

Exercises
Question 1.

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?


Answer:

Given,


Size of the candle, h = 2.5 cm


Object distance, u = –27 cm


Image size = h’ = ?


Radius of curvature of the concave mirror,


R = – 36 cm


Now focal length f is given as f = R/2


∴ f = R/2 = -18 cm



The image distance can be obtained using the mirror equation,



Where,


f is the focal length


u is object distance and


v is the image distance


⇒ 


⇒ 


⇒ 


⇒ v = -54 cm


Therefore, the screen should be placed at a distance of 54 cm


from the mirror to obtain a sharp image.


The magnification is given by



Where,


m is the magnification


h’ is the height of the image


h is the height of the object


v is the image distance


u is the object distance


Using the equation, we get,



⇒ 


⇒ h = -5 cm


∴ The height of the image is 5cm. The negative sign indicate that the image is real, inverted and magnified.




Question 2.

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.


Answer:

Given,


Size of the needle, h = 4.5 cm


Object distance, u = – 12 cm


Focal length of the convex mirror, f = 15 cm


The image distance can be obtained using the mirror equation,



Where,


f is the focal length


u is object distance and


v is the image distance


⇒ 


⇒ 


⇒ 


⇒ v = 6.67 cm


Therefore, the image of the needle is 6.67 cm away


from the mirror.


The magnification is given by



Where,


m is the magnification


h’ is the height of the image


h is the height of the object


v is the image distance


u is the object distance


Using the equation, we get,



⇒ 


⇒ h' = + 2.5 cm


⇒ 


⇒ m = 0.56


∴ Magnification is 0.56. And, the height of the image is 2.5 cm. The image is virtual.


If the needle is moved farther from the mirror the image will also move farther away. The size of the image will also decrease gradually.



Question 3.

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?


Answer:

Actual depth of the needle in water, h1 = 12.5 cm


Apparent depth of the needle in water, h2 = 9.4 cm


Refractive index of water = μ


The value of μ can be obtained as follows:


According to Snell’s law,



∴ μ = 1.33


Hence, the refractive index of water is about 1.33.


When the water is replaced by a liquid of refractive index,


 = 1.63


The actual depth of the needle remains the same, but the apparent depth of the needle changes.


EXPLANATION: This is because the light ray coming from the needle inside liquid suffers refraction at the liquid-air interface, as the refractive index of liquid is greater than air the light ray bends away from the normal at the point of incidence due to which the refracted ray appears to be coming from a point at a depth less than the actual depth of the needle.


Let h2 be the new apparent depth of the needle.


Hence, we can write the relation:




⇒ 


⇒ h2’ = 7.67 cm


Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2.


∴ Distance by which the microscope should move = 9.4 cm - 7.6 cm


= 1.73 cm upwards.



Question 4.

Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45 � with the normal to a water-glass interface [Fig. 9.34(c)].



Answer:

For the glass -air interface:



Angle of incidence, i = 60°


Angle of refraction, r = 35°


According to Snell’s law,



⇒ 


⇒ 


⇒ 


Note: refractive index is a unitless quantity.


For the water -air interface:



Angle of incidence, i = 60°


Angle of refraction, r = 47°


According to Snell’s law,



⇒ 


⇒ 


⇒ 


⇒ 


For the glass - water interface:



Angle of incidence, i = 450


Angle of refraction, r = ? 0


According to Snell’s law,



⇒ 


⇒ 



⇒ r ≈ 37°


Therefore, the angle of refraction in glass is 



Question 5.

A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)


Answer:

Given Data,


Depth of bulb in water = d = 80 cm = 0.8 m


Refractive index of water = μ = 1.33


Construction:


Please consider point “N” at the bottom of the tank where the bulb is placed.


Let the axis line “MOP” be the line on the surface of the water which is 80 cm from the bottom of the tank.


The below is the figure which we would get once we finish our construction.



From the figure we get,


Angle of incidence = Angle of reflection = 90°


Principle: The law of reflection governs the reflection of light-rays off smooth conducting surfaces (here water). The law of reflection states that the incident ray, the reflected ray, and the normal to the surface of the water all lie in the same plane. Furthermore, the angle of reflection is equal to the angle of incidence. Both angles are measured with respect to the normal to the water in the tank.


Since we consider bulb as a point source, the emergent light can be considered as a circle.


The radius of this circle can be obtained by,


R = MP/2 (MP is taken as diameter here)


= MO = MP (from the above figure)


According to Snell’s law, ratio of sine of angle of incidence and sine of angle of refraction is always constant for a given pair of media.


μ = sin r/ sin i


⇒  (substituting μ from given data & r from figure)


⇒ i = sin-1 (1/1.33) = 48.75° (from log tables)


From the figure,


tan i = OP/ON = R/d


⇒ R = d × tan(i)


⇒ R = tan 48.75° x 0.8 = 0.91 m


Area of the surface of water (since we considered the point source is bulb and the surface covered is a circle) = πR2 = π(0.91)2 = 2.61 m2


Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2



Question 6.

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.


Answer:

Given Data,


Angle of minimum deviation is δm = 40°.


Angle of the prism is 𝐴 = 60°.


Refractive index of water is μw = 1.33.


Refractive index of the material of the prism be μp .


The angle of deviation is related to refractive index as:


 ..(1)


Reference Figure for the above formula:



Explanation: There is an angle of incidence at which the sum of the two deflections is a minimum. The deviation angle at this point is called the "minimum deviation" angle, or "angle of minimum deviation". At the minimum deviation angle, the incidence and exit angles of the ray are identical. One of the factors that causes a rainbow is the bunching of light rays at the minimum deviation angle that is close to the rainbow angle.


Substituting the given data in the equation (1):



⇒ 


⇒ 


⇒ 𝜇′ = 1.532


Hence, we can conclude that the refractive index of the material of the prism is 1.532.


B. Since the prism is placed in water, let 𝛿′𝑚 be the new angle of minimum deviation for the same prism.


The refractive index of glass with respect to water is given by the relation:



Also, 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ (A + 𝛿′𝑚) = 2 × 35.160


⇒ (A + 𝛿′𝑚) = 70.320


⇒ 𝛿′𝑚 = 70.320 – 600(∵ A = 600)


∴ 𝛿′𝑚 = 10.320


Hence the new angle of minimum deviation of a parallel beam of light is 10.320



Question 7.

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?


Answer:

Given Data:


Refractive index of glass, μ = 1.55


Focal length of the double-convex lens, f = 20 cm


Consider,


Radius of curvature of one face of the lens = R1


Radius of curvature of the other face of the lens = R2


Radius of curvature of the double-convex lens = R


Since it is mentioned that both faces are of the same radius,


⇒ R1 = R and R2 = -R


The value of R can be calculated as: t


… (1)


(Explanation: The above formula is generated by the ray diagram of the double convex lens. )


Substituting the given data in the equation (1), we get,




⇒ R = 0.55 × 2 × 20


⇒ R = 22 cm


Hence, the radius of curvature of the double-convex lens is 22 cm.



Question 8.

A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?


Answer:

Given data,


The Lens is placed in the path of the convergent beam 12 cm from P.


Here, the object is virtual and the image is real, u = + 12 cm (since, the object on right and virtual)


Reference ray diagram:



a) Given f = + 20 cm for the convex lens.


Using Len’s formula:





⇒ 


⇒ 


i.e. v = 7.5 cm (image on right and real)


This means that the beam converges as a point 7.5 cms from the lens


b) Here, f = -16 cm


According to the lens formula, we have the relation:


l



⇒ 


⇒ 


⇒ v = 48 cm


Hence, the image will be located 48 cm from the lens.



Question 9.

An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?


Answer:

Given data,

Size of the object, h1 = 3 cm


Object distance, u = – 14 cm


Focal length of the concave lens, f = – 21 cm


Image distance = v


According to the lens formula, we have the relation:



⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ v = - 8.4 cm


Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.


The magnification of the image is given as:




⇒ h2 = 0.6 × 3 = 1.8 cm


Hence, the height of the image is 1.8 cm.


If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.



Question 10.

What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.


Answer:

Given Data,

Focal length of the convex lens, f1 = 30 cm


Focal length of the concave lens, f2 = – 20 cm


Focal length of the system of lenses = f


The equivalent focal length of a system of two lenses in contact is given as:



⇒ 


⇒ 


⇒ f = -60 cm


Hence, the focal length of the combination of lenses is 60 cm.


The negative sign indicates that the system of lenses acts as a diverging lens.



Question 11.

A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?


Answer:

Given:


Focal length of the objective lens, f1 = 2.0 cm


Focal length of the eyepiece lens, f2 = 6.25 cm


Distance between eyepiece and objective, d = 15 cm



We need to find the distance of object from objective lens for the following two cases:


(a) The final image is formed at least distance of distinct vision, v2 = 25 cm


Let the distance of object from eyepiece lens = u2 cm


By using lens maker formula,


 …(1)


From equation 1 we have,


…(2)


Where, f2 = focal length of the eye piece lens


v2 = Distance of image formation


u2 = Distance of object from eyepiece


Plugging the values in equation (2)



⇒ 


⇒ u2 = -5 cm


We understand that image distance from objective lens v1, is given by


v1 = d + u2


v1 = 15cm - 5cm


v1 = 10 cm


Let image distance for the objective lens = u1


From equation (1), for objective lens, we have,


…(3)


Plugging the values in equation (3) we have,



⇒ u1 = -2.5 cm


We know that magnifying power of a compound lens is given by,


m = ) …(4)


Where,


m = magnification


v1 = image distance from objective lens.


u1 = object distance from objective lens


d = minimum distance of distinct vision (25 cm)


f2 = Focal length of eyepiece


Now putting the values in equation (4)



m = 20.


Hence the magnification of the compound microscope in this case is 20. Magnification is a unit less quantity, as you can observe from the equation.


(b) Distance of final image formed, v2 = ∞


Object distance for the eyepiece = u2


By using lens maker formula we have,


 …(1)


From equation 1 we have,


…(2)


Where, f2 = focal length of the eye piece lens


v2 = Distance of image formation


u2 = Distance of object from eyepiece


Plugging the values in equation (2)


We get,



u2 = -6.25 cm


We understand that image distance from objective lens v1, is given by


v1 = d + u2


v1 = 15cm – 6.25cm


v1 = 8.75 cm


Let image distance for the objective lens = u1


From equation (1), for objective lens, we have,


…(3)


Plugging the values in equation (3) we have,



⇒ 


⇒ u1 = -2.59 cm


We know that magnifying power of a compound lens is given by,


m =  …(4)


Where,


m = magnification


v1 = image distance from objective lens.


u1 = object distance from objective lens


d = minimum distance of distinct vision (25 cm)


f2 = Focal length of eyepiece


Now putting the values in equation (4)



m = 13.51


Hence, the magnifying power of the microscope in this case is 13.51.



Question 12.

A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.


Answer:

Given:


Focal length of the objective lens, f0 = 8 mm (0.8 cm)


Focal length of the eyepiece, fe = 2.5 cm


Object distance for the objective lens, u0 = -9.0 mm ( -0.9 cm)


Least distance of distant vision, d = 25 cm


Image distance for the eyepiece, ve = -d = -25 cm


Let object distance for the eyepiece = ue.



By using lens maker formula,


 …(1)


From equation 1 we have,


…(2)


Where, fe = focal length of the eye piece lens


ve = Distance of image formation


ue = Distance of object from eyepiece


Plugging the values in equation (2)



⇒ 


⇒ u2 = -2.27 cm


We can find out the distance of image for objective lens using the equation (1)


For objective lens, we have


From equation 1 we have,


…(3)


Where, f0 = focal length of the objective lens


v0 = Distance of image formation


u0 = Distance of object from objective


Plugging the values in equation (3)



⇒ v0 = 7.2 cm


We know that the distance between the objective lens and eyepiece, d is given by,


d = |ue| + vo


d = 2.72 + 7.2


d = 9.47 cm


We know that magnifying power of a compound lens is given by,


m = ) …(4)


Where,


m = magnification


v0 = image distance from objective lens.


U0 = object distance from objective lens


d = minimum distance of distinct vision (25 cm)


fe = Focal length of eyepiece


Now putting the values in equation (4)


m = )


m = 88.


Hence the magnification power of the microscope is 88 and it is a dimensionless quantity as obvious from calculation.



Question 13.

A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?


Answer:

Given:


Focal length of the objective lens, fo = 144 cm
Focal length of the eyepiece, fe = 6.0 cm



The magnifying power of the telescope is given as,


 …(1)


Where,


f0 = focal length of objective lens


fe = focal length of eyepiece


By putting the values in equation (1) we get,


⇒ m = 144/6


⇒ m = 24


The separation between the objective lens and the eyepiece is calculated as:


d = fo + fe


⇒ d = 144 + 6


⇒ d = 150cm


Hence 24 is the magnifying power of the telescope and the separation between the objective lens and the eyepiece is 150cm.


NOTE: Separation of lenses or length of telescope in case of refractive telescopes is always equal to the sum of focal lengths of the two lenses because one lens converges the rays from infinity at its focus and the eyepiece magnifies the image formed( virtual object) at its focus.



Question 14.

(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm is used, what is the angular magnification of the telescope?

(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106m, and the radius of lunar orbit is 3.8 × 108m.


Answer:

Given:


Focal length of the objective lens, fo = 15m = 1500 cm
Focal length of the eyepiece, fe = 1.0 cm



(a) The angular magnification of a telescope is given as:


α = fo × fe


⇒ α = 1500 × 1


⇒ α = 1500
Hence, the angular magnification of the given refracting telescope is 1500.


(b) Diameter of the moon, d = 3.48 × 106 m
Radius of the lunar orbit, r0 = 3.8 × 108 m
Let d’ be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.


The relation between the given variables is,


 …(2)


Where,


d = radius of the moon


r0 = radius of lunar orbit or the distance of moon from earth.


d' = 


d′ = 13.74 cm


Hence the diameter of the image of moon formed by telescope is 13.74 cm.



Question 15.

Use the mirror equation to deduce that:

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

(b) a convex mirror always produces a virtual image independent of the location of the object.

(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]


Answer:

Given:


(a) For a concave mirror, the focal length (f) is negative.
Therefore f < 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
For image distance v, we can write the lens formula as:


 …(1)


The object lies between f and 2f.


2f < u < f


⇒ 1/2f < 1/u < 1/f


⇒ −1/2f < −1/u < −1/f


0 …(ii)


Using equation (1), we get:


1/2f < 1/v < 0


1/v is negative, i.e., v is negative.


1/2f < 1v


2f > v


-v > -2f
Therefore, the image lies beyond 2f.


(b) For a convex mirror, the focal length (f) is positive.
Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
For image distance v, we have the mirror formula:


 …(3)


Using equation (2), we can conclude that:


1/v < 0


v > 0


Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, irrespective of the object distance.


(c) For a convex mirror, the focal length (f) is positive.
Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative,
Therefore u < 0
For image distance v, we have the mirror formula:



⇒ 


But we have u < 0


Therefore 1/v > 1/f


v < f
Hence, the image formed is diminished and is located between the focus (f) and the pole.


(d) For a concave mirror, the focal length (f) is negative.


Therefore f < 0


When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
It is placed between the focus (f) and the pole.


Therefore f > u > 0


1/f < 1/u < 0


1/f − 1/u < 0
For image distance v, we have the mirror formula:



Therefore 1/v < 0


v > 0
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:


1/u > 1/v


v > u
Magnification, m = v/u > 1
Hence, the formed image is enlarged



Question 16.

A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?


Answer:

Given:


Actual depth of the pin, d = 15 cm
Let apparent depth of the pin = d’ cm
Refractive index of glass, μ = 1.5



Ratio of actual depth to the apparent depth is equal to the refractive index of glass,
i.e.


μ = d / d′


⇒ d’ = d × μ


⇒ d = 15 × 1.5 = 10cm


The distance at which the pin appears to be raised D,


D = d’-d


D = 15-10 = 5cm


So the pin appears to be raised by 5 cm.
For a small angle of incidence, this distance does not depend upon the location of the slab.



Question 17.

(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.

(b) What is the answer if there is no outer covering of the pipe?



Answer:

Given:


(a) Refractive index of the glass fibre,


Refractive index of the outer covering of the pipe, μ2 = 1.44
Angle of incidence = i°
Angle of refraction = r°
Angle of incidence at the interface = i
The refractive index (μ) of the core interface is given as:


μ =  …(1)


Where,


2 = refractive index of outer covering


1 = refractive index of inner covering


i = angle of incidence at interface


From equation (1)


sin i = �1/ �2


⇒ sin i = 1.44/1.68


⇒ i' = 59°



For the critical angle, total internal reflection takes place only when i > i’,


i.e., i > 59°
Maximum angle of reflection,


rmax = 90°−i′


⇒ rmax = 90°−59° = 31°
Let, imax be the maximum angle of incidence.
The refractive index at the air − glass interface, μ1 = 1.68


⇒ 


⇒ sin imax = �1 × sin rmax


sin imax = 1.68 × 0.5150 = 0.8652


Therefore imax = sin-1(0.86)


imax = 60°


Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.


(b) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, μ1 = Refractive index of air = 1
For the angle of incidence i = 90°, we can write Snell’s law at the air − pipe interface as:
sin i /sin r = μ


Where,


i = angle of incidence


r = angle of refraction


sin i/ sin r = 1.68


sinr = sin90°/1.68


Sin r = 1/1.68


r = 36.5°


i = 90° - r


⇒ i′ = 90°−36.5°


⇒ i = 53.5°


Since i’ > r, all incident rays will suffer total internal reflection.



Question 18.

Answer the following question:

You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.


Answer:

Yes,


Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.



Question 19.

Answer the following question:

A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?


Answer:

No,


When light rays diverge, a virtual image is formed. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.




Question 20.

Answer the following question:

A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?


Answer:

The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result the fisherman will appear to be taller.




Question 21.

Answer the following question:

Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?


Answer:

Yes, decrease, the apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.



Question 22.

Answer the following question:

The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?


Answer:

Yes,


The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces, so that the diamond sparkles.



Question 23.

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?


Answer:

Given:


Distance between the object and the image, d = 3 m



Maximum focal length of the convex lens = fmax
For real images, the maximum focal length is given by,


fmax = d/4 …(1)


Where,


fmax = maximum focal length


d = distance between object and image


By putting the values in equation (1) , we get,


fmax = 0.75 m


Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.



Question 24.

A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Determine the focal length of the lens.


Answer:

Given:


Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:


f =  …(1)


Where,


f = focal length


D = distance between image and the object.


d = Distance between two locations of the convex lens.


By putting the values in equation (1) we get,



⇒ f = 21.39cm
Therefore, the focal length of the convex lens is 21.39 cm.



Question 25.

Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?


Answer:

Given:


Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = −20 cm
Distance between the two lenses, d = 8.0 cm
When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:


 …(1)


Where,
u1 = Object distance
v1 = Image distance


f1 = Focal length of convex lens.


By putting the values in equation (1), we get,



By solving the above equation we get,


v1 = 30cm
This image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:


 …(2)


Where,
u2 = Object distance


u2 = image distance- distance between lenses
⇒ u2 = (30 − d) = 30 − 8 = 22 cm


v2 = image distance


f2 = focal length of concave lens


Now, putting the values in equation (2) we get,


⇒ 


Therefore v2 = −220cm


The parallel incident beam appears to diverge from a point that
is (220−d2) = (220−4) = 216cm from the centre of the combination of the two lenses.
When the parallel beam of light is incident, from the left, on the concave lens I)
According to the lens formula, we have:


 …(3)


From equation (3) we get,



Where,
u2 = Object distance = −∞


v2 = Image distance


by putting the values in above equation we get,



Therefore v2 = −20cm


The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:


 …(4)


Where,
u1 = Object distance
⇒ u1 = −(20 + d) = −(20 + 8) = −28 cm
v1 = Image distance




Therefore v2 = −420cm


Hence, the parallel incident beam appear to diverge from a point that is (420 − 4) = 416 cm from the left of the centre of the combination of the two lenses. The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.



Question 26.

An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm. Determine the magnification produced by the two-lens system, and the size of the image.


Answer:

Given:


Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = −20 cm
Distance between the two lenses, d = 8.0 cm
Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens, u1 = -40 cm
|u1| = 40cm


According to the lens formula:


 …(1)


Where,
u1 = Object distance
v1 = Image distance


f1 = Focal length of convex lens.


By putting the values in equation (1), we get,



By solving the above equation we get,


v1 = 120 cm
Magnification, m = |v1|/|u1| = 120/40 = 3
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.


According to the lens formula:


 …(2)


Where,
u2 = Object distance


U2 = + (120 − 8) = 112 cm.
v2 = Image distance


f2 = Focal length of convex lens.


By putting the values in equation (2), we get,



By solving the above equation, we get,


v2 = -2240/92


magnification, m’ = |v2/u2| = 2240/92 × 1/112 = 20/92


Hence, the magnification due to the concave lens is 2092.
The magnification produced by the combination of the two lenses is calculated as:


M = m × m = 0.652
The magnification of the combination is given as:


h2/h1 = 0.652


h2 = h1 × 0.652
Where,
h1 = Object size = 1.5 cm
h2 = Size of the image
Therefore h2 = 0.652 × 1.5 = 0.98cm


Hence, the height of the image is 0.98 cm.



Question 27.

At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.


Answer:

Given:


Angle of prism, A = 60°
Refractive index of the prism, μ = 1.524


Emergent angle, e = 90°



i1 = Incident angle
r1 = Refracted angle
r2 = Angle of incidence at the face AC
According to Snell’s law, for face AC, we can have:


 …(1)


From equation(1) we have,


⇒ sin r2 = 


Putting the values in the above equation we find that,


r2 = sin


r2 = 41°


We know that, refracted angle r1 is given by,


r1 = A- r2


r1 = 60-41 = 19°


Therefore r1 = A−r2 = 60−41 = 19°
According to Snell’s law, we have the relation:


μ = 


Sin i1 = 1.524 × sin19° = 0.496


⇒ i1 = 29.75°


Therefore, the angle of incidence is 29.75°



Question 28.

You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will

(a) deviate a pencil of white light without much dispersion,

(b) disperse (and displace) a pencil of white light without much deviation.


Answer:

(a) Place the two prisms beside each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism, it will recombine and white light will emerge from the combination of the two prisms.
(b) Take the system of the two prisms as suggested in answer (a). Increase the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will disperse the pencil of white light without much deviation.



Question 29.

For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.


Answer:

Given:


Least distance of distinct vision, d = 25 cm
Far point of a normal eye, d’ = ∞
Converging power of the cornea,
Least converging power of the eye-lens, Pc = 40D
To see the objects at infinity, the eye uses its least converging power.
Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60D


Power of the eye-lens is given as:


P = 1/ focal length


f = 1/P × 1/60D


100/60 = 5/3cm
To focus an object at the near point, object distance (u) = −d = −25 cm
Focal length of the eye-lens = Distance between the cornea and the retina = Image
distance
Hence, image distance, v = 53
Applying the lens formula we have:


…(1)


Where, f0 = focal length of the objective lens


v0 = Distance of image formation


u0 = Distance of object from objective


1/f = 3/5 + 1/25


1/f = (15 + 1)/25 = 16/25cm-1


Power P’ = 1/f′ × 100


P = 16/25 × 100 = 64D


Therefore Power of the eye-lens = 64 − 40 = 24 D
Hence, the range of accommodation of the eye-lens is from 20D to 24D



Question 30.

Does short-sightedness (myopia) or long-sightedness (hyper - metropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?


Answer:

Given:


When the eye-lens loses its ability of accommodation, the defect is called presbyopia. In some cases, the eye-lens is normally accommodated by a person undergoing through myopia or hypermetropia. When the eye-balls get elongated from front to back then it is myopia. Hypermetropia, on the other hand occurs when the eye-balls get shortened.



Question 31.

A myopic person has been using spectacles of power –1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.


Answer:

The myopic person uses a spectacle of power, P = -1.0 D


Focal length of the given spectacles, f = 1/P


f = 1/(-1 × 10-2 )


f = -100 cm


Hence, 100 cm is the far point of the person. He might have a normal near point of 25 cm. The objects placed at infinity produce virtual images at 100 cm, when spectacles are being used. He is able to see the objects placed between 100 cm and 25 cm when he uses the ability of accommodation of the eye-lens.


During old age, the person uses reading glasses of power, P’ = + 2 D


Old age decreases the ability of accommodation. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.



Question 32.

A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?


Answer:

The person is not able to see horizontal lines clearly, whereas the vertical lines are distinctly visible. This problem arises due to problem in refracting mechanism of eye specially cornea and eye lens. This condition is called as astigmatism. The curvature of eye lens in the horizontal plane may be abnormal. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.



Question 33.

A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?


Answer:

Given:


(a) Focal length of the magnifying glass, f = 5 cm


Distance vision has the least distance, d = 25 cm


Closest object distance = u


Image distance, v = -d = -25 cm


Applying the lens formula we have:


…(1)


Where, f0 = focal length of the objective lens


v0 = Distance of image formation


u0 = Distance of object from objective



⇒ u = -25/6


⇒ u = 4.16 cm


Hence, the closest distance at which the person can read the book is 4.167 cm.


For the object at the longest distance (u’), the image distance (v’) = ∞


Again using equation (1), we get,



By solving the above equation we get,


u = – 1/5


Therefore, u’ = -5 cm


Hence, the farthest distance at which the person can read the book is 5 cm.


(b) Maximum angular magnification is given by the relation:


αmax = d/|u′|


= 25/5


= 5


Hence the maximum magnification is 5.



Question 34.

A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.


Answer:

Given:


Area of each square, A = 1 mm2


Object distance, u = −9 cm


Focal length of a converging lens, f = 10 cm


(a) Applying the lens formula we have:


…(1)


Where, f0 = focal length of the objective lens


v0 = Distance of image formation


u0 = Distance of object from objective


The lens formula for the image distance v, can be written as:


…(2)


By putting available values in equation (2) we get,



⇒ V0 = -90 cm


Magnification, m = v/u


m = −90/−9


m = 10


Therefore, area of each square in the virtual image = (10)2 A


= 102 x 1 = 100 mm2


= 1 cm2


(b) The lens has a magnifying power of = d/ |u|


= 25/9


= 2.8


(c) The magnification in (a) is not the same as the magnifying power in (b). Both the quantities will be same once the image is formed at the near point (25 cm)



Question 35.

At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?


Answer:

Given:


Image distance, v = −d = −25 cm


Focal length, f = 10 cm


Object distance = u


When the image is formed at the near point or at the minimum distance of distinct vision, we get the maximum possible magnification.


Applying the lens formula we have:


…(1)


Where, f0 = focal length of the objective lens


v0 = Distance of image formation


u0 = Distance of object from objective


Putting values in the equation (1) we get,


⇒ 


⇒ u = -50/7


⇒ u = -7.14 cm


Hence, the lens should be kept 7.14 cm away in order to view the squares distinctly.



Question 36.

What is the magnification in this case?


Answer:

Magnification = |v/u|


= 25/7.14 = 3.5



Question 37.

Is the magnification equal to the magnifying power in this case? Explain.


Answer:

Magnifying power = d/u


= 25/7.14


= 3.5


Therefore, the magnifying power is equal to the magnitude of magnification since the image is formed at the near point (25 cm).



Question 38.

What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?


Answer:

Given:


Area of the virtual image of each square, A = 6.25 mm2


Area of each square, A0 = 1 mm2


Hence, the linear magnification of the object can be calculated as:


 …(1)


Where, m = magnification


A = Area of virtual image


A0 = Area of each square


Putting values in equation (1) we get,


m = 


⇒ m = 2.5


Also,


m = Image distance(v)/Object distance(u)


m = v/u


Therefore, v = m × u


= 2.5u …(2)


The magnifying glass has a focal length of, f = 10 cm


Applying the lens formula for lens we have:


…(3)


Where, f0 = focal length of the objective lens


v0 = Distance of image formation


u0 = Distance of object from objective


Putting values in the equation (3) we get,


⇒ 


⇒ u = -(1.5 × 10) / 2.5


Therefore, u = -6 cm


And v = 2.5u


⇒ v = 2.5 × 6


⇒ v = -15 cm


The virtual image cannot be seen by the eyes distinctly, because the image is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye.



Question 39.

Answer the following question:

The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?


Answer:

The angular size of the image is equal to the angular size of the object, but still the angular size of image is equal to angular size of the object. The objects placed closer than the least distance of distinct vision (i.e., 25 cm) can be viewed through a magnifying glass. A closer object causes a larger angular size. A magnifying glass provides angular magnification. The object cannot be placed closer to the eye, without magnification. With magnification, the object can be placed much closer to the eye.



Question 40.

Answer the following question:

In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?


Answer:

Yes, the angular magnification changes. The angular magnification decreases a little, when the distance between the eye and a magnifying glass is increased. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.



Question 41.

Answer the following question:

Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?


Answer:

Making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length. Thus, the focal length of a convex lens cannot be decreased by a greater amount.



Question 42.

Answer the following question:

Why must both the objective and the eyepiece of a compound microscope have short focal lengths?


Answer:

A compound microscope produces an angular magnification of [(25/fe) + 1]


where,


fe = focal length of the eyepiece


It can be inferred that if fe is small, then angular magnification of the eyepiece will be large.


The angular magnification of the objective lens of a compound microscope is given as


1/(|u0|f0)


Where, uo = object distance for the objective lens


fo = focal length of the objective


The magnification is large when uo > fo. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since, uo is small, fo will be even smaller. Therefore, fe and fo are both small in the given condition.



Question 43.

Answer the following question:

When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?


Answer:

We are unable to collect much refracted light when we place our eyes too close to the eyepiece of a compound microscope. As a result, there is substantial decrement in the field of view. Hence, the clarity of the image gets blurred. The eye-ring attached to the eyepiece gives the best position for viewing through a compound microscope. The precise location of the eye depends on the separation between the objective lens and the eyepiece.



Question 44.

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?


Answer:

Given:


Focal length of the objective lens, = 1.25 cm


Focal length of the eyepiece, fe = 5 cm


Least distance of distinct vision, d = 25 cm


Angular magnification of the compound microscope = 30X


Total magnifying power of the compound microscope, m = 30


The angular magnification of the eyepiece is given by the relation:


 …(1)


By putting the values in equation 1, we get


⇒ m = (1 + 25/5)


⇒ m = 6


The angular magnification of the objective lens (m0) is related to me as:


mo × me = m


⇒ mo = m/ me


⇒ m0 = 30/6


⇒ m0 = 5


We also have the relation:



From the available data we find that,


5 = v0/(-u0)


Therefore, vo = -5uo ………. (2)


Applying the lens formula for the objective lens:


…(3)


Where, f0 = focal length of the objective lens


v0 = Distance of image formation


u0 = Distance of object from objective


Plugging the values in equation (3)



⇒ u0 = (-6/5) × 1.25


⇒ u0 = -1.5 cm


By putting the value of u0 in equation (2) we can find v0,


v0 = -5 × -1.5


v0 = 7.5 cm


1fo = 1vo–1uo 11.25 = 1−5uo−1uo


= −65uo


Therefore, uo = −65 × 1.25


= -1.5 cm


And vo = -5uo


= -5 x (-1.5)


= 7.5 cm


The object should be placed 1.5 cm away from the objective lens to obtain the required magnification.


Applying the lens formula for the eyepiece:


Applying the lens formula for the objective lens:


…(3)


Where, fe = focal length of the objective lens


ve = Image distance for eyepiece = -25cm


u0 = Distance of object from objective


Plugging the values in equation (3)



⇒ ue = (-6/25)


⇒ ue = -4.17 cm


Separation between the objective lens and the eyepiece d = |ue| + |vo|


d = 4.17 + 7.5


d = 11.67 cm


Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.


NOTE: Separation of lenses or length of telescope in case of refractive telescopes is always equal to the sum of focal lengths of the two lenses because one lens converges the rays from infinity at its focus and the eyepiece magnifies the image formed( virtual object) at its focus.



Question 45.

A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?

(b) the final image is formed at the least distance of distinct vision (25cm)?


Answer:

Given:


Focal length of the objective lens, = 140 cm


Focal length of the eyepiece, fe = 5 cm


Least distance of distinct vision, d = 25 cm


(a) When the telescope is in normal adjustment, its magnifying power is given as:


m = fo/fe …(1)


By putting the given values in equation (1), we get,


m = 140/5


⇒ m = 28


(b) When the final image is formed at d, the magnifying power m, of the telescope is given as:


 …(2)


By putting the values in equation (2), we get,



⇒ m = 28 × (1 + 0.2)


⇒ m = 33.6


Hence, the magnification in case (a) is 28 and in case (b) is 33.6.



Question 46.

For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?


Answer:

Given:


Focal length of the objective lens, f0 = 140 cm


Focal length of the eyepiece, fe = 5 cm


In normal adjustment, the separation between the objective lens and the eyepiece


(The rays striking the objective lens are coming from infinity hence they are converged at the focus by objective lens, a small image is formed at focus of objective, the eyepiece magnifies the image at it focus, hence the length of telescope or distance between the objective and eyepiece is fe + fo)


d = fo + fe


⇒ d = 140 + 5


⇒ d = 145 cm



Question 47.

If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?


Answer:

Given:


Focal length of the objective lens, f0 = 140 cm


Focal length of the eyepiece, fe = 5 cm


Height of the tower, h1 = 100 m


Distance of the tower (object) from the telescope, u = 3Km = 3000 m


The angle subtended by the tower at the telescope is given as:


θ = h1/ u


⇒ θ = 100/3000


⇒ θ = 1/30 rad


The angle subtended by the image produced by the objective lens is given as:


θ = h2/ fo


⇒ θ = h2 /140 rad


Where, h2 = height of the image of the tower formed by the objective lens


1/30 = h2 /140


Therefore,


h2 = 140/30


⇒ h2 = 4.7 cm


Therefore, the objective lens forms a 4.7 cm tall image of the tower.



Question 48.

What is the height of the final image of the tower if it is formed at 25cm?


Answer:

Given:


Focal length of the objective lens, f0 = 140 cm


Focal length of the eyepiece, fe = 5 cm


Image is formed at a distance, d = 25cm


The magnification of the eyepiece is given by the relation:


m = 1 + d/fe


⇒ m = 1 + 25/5


⇒ m = 6


Height of the final image H, is given by,


H = m × h2 = 6 × 4.7 = 28.2 cm


Hence, the height of the final image of the tower is 28.2 cm.



Question 49.

A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?


Answer:

Given:


Distance between the objective mirror and the secondary mirror, d = 20 mm


Radius of curvature of the objective mirror, R1 = 220 mm


Therefore, focal length of the objective mirror, f1 = 0.5 × R1 = 110 mm


Radius of curvature of the secondary mirror, R2 = 140 mm


Therefore, focal length of the secondary mirror, f2 = 0.5 × R2 = 70 mm


The image of an object placed at infinity, formed by the objective mirror, acts as a virtual object for the secondary mirror.


Let the virtual object distance for the secondary mirror be u,


u = f1 – d …(1)


u = 110 – 20


u = 90 mm


By applying the mirror formula for the secondary mirror, we can calculate image distance (v) as:


 …(2)


By putting the values in equation (2), we get,



v = 315 mm


Hence, the final image will be formed 315 mm away from the secondary mirror.


NOTE: Cassegrainian telescopes are one of the most widely used telescope of current times. It is a combination of a parabolic primary mirror and secondary hyperbolic mirror.



Question 50.

Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. A current in the coil produces a deflection of 3.50 of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?



Answer:

Given:


Angle of deflection, θ = 3.5°


Distance of the screen from the mirror, D = 1.5 m


The reflected rays get deflected by an amount twice the angle of deflection i.e.,


2θ = 7.0°


The displacement (d) of the reflected spot of light on the screen is given as:


tan2θ = d/1.5


(Using trigonometric ratio)


d = 1.5 x tan7o


d= 0.184 m = 18.4 cm


Hence, the displacement of the reflected spot of light is 18.4 cm



Question 51.

Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?



Answer:

Given:


Focal length of the convex lens, f1 = 30 cm


The liquid acts as a mirror. Focal length of the liquid = f2


Focal length of the system (convex lens + liquid), f = 45 cm


Here we have combination of lenses,


For a pair of optical systems placed in contact, the equivalent focal length is given as:


 …(1)


By putting the values in equation (1), we can evaluate f2,



f2 = – 90cm


Hence, the focal length of liquid is -90 cm


Let the refractive index of the lens be and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is −R.


R can be obtained using the relation:


 = (μ1−1)() …(2)


Putting the values in equation (2), we get,


130 = (1.5 – 1) (2/R)


By the solving the above equation we get,


R = 30cm


Let μ2 be the refractive index of the liquid.


Radius of curvature of the liquid on the side of the plane mirror = ∞


Radius of curvature of the liquid on the side of the lens, R = −30 cm


The value of μ2 can be calculated using the relation:


 = (μ2−1) × () …(3)


By plugging the values in equation 3, we get,


μ2 – 1 = 1/3


Therefore, μ2 = 1.33


Hence, the refractive index of the liquid is 1.33.


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