Solved Problems in Optics: Class 10 Science Chapter 2 Guide

Solved Problems: Optics - Science

This study guide provides detailed solutions to key problems from Chapter 2: Optics, covering essential concepts for Class 10 Science.

Problem 1

Light rays travel from vacuum into a glass whose refractive index is 1.5. If the angle of incidence is 30°, calculate the angle of refraction inside the glass.

Solution:

According to Snell’s law,

$$\frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1}$$

This can be written as:

$$\mu_1 \sin i = \mu_2 \sin r$$

Here, the light travels from vacuum (medium 1) to glass (medium 2). The given values are:

• Refractive index of vacuum, \(\mu_1\) = 1.0
• Refractive index of glass, \(\mu_2\) = 1.5
• Angle of incidence, \(i\) = 30°

Substituting the values into the equation:

\((1.0) \sin 30° = 1.5 \sin r\)

\(1 \times \frac{1}{2} = 1.5 \sin r\)

\(\sin r = \frac{1}{2 \times 1.5} = \frac{1}{3} \approx 0.333\)

Now, we find the angle of refraction, \(r\):

\(r = \sin^{-1}(0.333)\)

\(r = 19.45°\)

Problem-2

A beam of light passing through a diverging lens of focal length 0.3m appear to be focused at a distance 0.2m behind the lens. Find the position of the object.

Solution:

For a diverging (concave) lens, the focal length (f) and image distance (v) are taken as negative according to the sign convention.

• Focal length, \(f\) = −0.3 m
• Image distance, \(v\) = −0.2 m

We use the lens formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

Rearranging to solve for the object distance (u):

$$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$$

Substituting the given values:

The position of the object is 0.6 m in front of the lens.

Problem-3

A person with myopia can see objects placed at a distance of 4m. If he wants to see objects at a distance of 20m, what should be the focal length and power of the concave lens he must wear?

Solution:

Given that the person's far point is x = 4m and they want to see an object at a distance y = 20m. We need a concave lens that will form an image of the object at 20m at the person's far point of 4m.

The focal length of the correction lens is given by the formula (Refer eqn.2.7):

$$f = \frac{xy}{x-y}$$

Substituting the values:

The focal length of the concave lens required is -5 m.

Now, we calculate the power of the correction lens:

$$P = \frac{1}{f}$$

The power of the correction lens is -0.2 Dioptre (D).

Problem-4

For a person with hypermetropia, the near point has moved to 1.5m. Calculate the focal length of the correction lens in order to make his eyes normal.

Solution:

The goal is to use a convex lens that allows the person to see objects placed at the normal near point (D) by forming their image at the person's actual near point (d).

Given that:

• The person's near point, d = 1.5m
• The normal near point for a healthy eye, D = 25cm = 0.25m

From equation (2.8), the focal length of the correction lens is:

$$f = \frac{d \times D}{d - D}$$

Substituting the values:

$$f = \frac{1.5 \times 0.25}{1.5 - 0.25} = \frac{0.375}{1.25} = 0.3 \text{ m}$$

The focal length of the convex lens needed for correction is 0.3 m.

Understanding Common Eye Defects: Myopia, Hypermetropia, Presbyopia, and Astigmatism

Defects in Eye

DEFECTS IN EYE

A normal human eye can clearly see all the objects placed between 25cm and infinity. But, for some people, the eye loses its power of accommodation. This could happen due to many reasons including ageing. Hence, their vision becomes defective. Let us discuss some of the common defects of human eye.

Myopia

Myopia, also known as short sightedness, occurs due to the lengthening of eye ball. With this defect, nearby objects can be seen clearly but distant objects cannot be seen clearly. The focal length of eye lens is reduced or the distance between eye lens and retina increases. Hence, the far point will not be infinity for such eyes and the far point has come closer. Due to this, the image of distant objects are formed before the retina (Figure 2.16-a). This defect can be corrected using a concave lens (Figure 2.16-b). The focal length of the concave lens to be used is computed as follows:

Figure 2.16 (a) Vision with myopia (b) Corrected vision using a concave lens

Let a person with myopia eye can see up to a distance x. Suppose that he wants to see all objects farther than this distance, i.e., up to infinity. Then the focal length of the required concave lens is f = –x. If the person can see up to a distance x and he wants to see up to a distance y, then, the focal length of the required concave lens is,

\( f = \frac{xy}{x-y} \)

Hypermeteropia

Hypermeteropia, also known as long sightedness, occurs due to the shortening of eye ball. With this defect, distant objects can be seen clearly but nearby objects cannot be seen clearly. The focal length of eye lens is increased or the distance between eye lens and retina decreases. Hence, the near point will not be at 25cm for such eyes and the near point has moved farther. Due to this, the image of nearby objects are formed behind the retina (Figure 2.17-a). This defect can be corrected using a convex lens (Figure 2.17-b). The focal length of the convex lens to be used is computed as follows:

Figure 2.17 (a) Vision with hypermetropia (b) Corrected vision using a convex lens

Let a person with hypermeteropia eye can see object beyond a distance d. Suppose that he wants to see all objects closer than this distance up to a distance D. Then, the focal length of the required convex lens is

\( F = \frac{dD}{d-D} \)

Presbyopia

Due to ageing, ciliary muscles become weak and the eye-lens become rigid (inflexible) andso the eye loses its power of accommodation.

Because of this, an aged person cannot see the nearby objects clearly. So, it is also called as ‘old age hypermetropia’.

Some persons may have both the defects of vision - myopia as well as hypermetropia. This can be corrected by ‘bifocal lenses’. In which, upper part consists of concave lens (to correct myopia) used for distant vision and the lower part consists of convex lens (to correct hypermetropia) used for reading purposes.

Astigmatism

In this defect, eye cannot see parallel and horizontal lines clearly. It may be inherited or acquired. It is due to the imperfect structure of eye lens because of the development of cataract on the lens, ulceration of cornea, injury to the refracting surfaces, etc. Astigmatism can be corrected by using cylindrical lenses (Torrid lenses).

Key Differences Between Convex and Concave Lenses Explained

Differences between a Convex Lens and a Concave Lens

A convex lens is thicker in the middle than at edges. A concave lens is thinner in the middle than at edges.

Convex Lens

1. A convex lens is thicker in the middle than at edges.
2. It is a converging lens.
3. It produces mostly real images.
4. It is used to treat hypermeteropia.

Concave Lens

1. A concave lens is thinner in the middle than at edges.
2. It is a diverging lens.
3. It produces virtual images.
4. It is used to treat myopia.

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

10th Science : Chapter 2 : Optics : Differences between a Convex Lens and a Concave Lens

Applications of Convex Lenses: Uses in Cameras, Microscopes, and Vision Correction

Applications of Convex Lenses

10th Science | Chapter 2: Optics

Key Applications of Convex Lenses

1. Convex lenses are used as camera lenses
2. They are used as magnifying lenses
3. They are used in making microscope, telescope and slide projectors
4. They are used to correct the defect of vision called hypermetropia

Reference Keywords: Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail