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.

Class 10 Science Chapter 2: Key Points to Remember on Optics

Optics: Key Points to Remember

Chapter 2: Optics Summary

• Light is a form of energy which travels along a straight line.
• The deviation in the path of light ray is called refraction.
• The ratio of speed of light in vacuum to the speed of light in a medium is defined as refractive index ‘µ’ of that medium.
• Lens formula:
• Magnification (m) is given by \( m = \frac{h'}{h} = \frac{v}{u} \).
• Power of lens, \( P = \frac{1}{f} \).
• The ability of the eye to focus nearby as well as the distant objects is called power of accommodation of the eye.
• A microscope is an optical instrument which helps us to see the objects which are very small in dimension.
• Telescope is an optical instrument used to see the distant objects clearly.

Understanding Refraction Through a Concave Lens: Image Formation Explained

Refraction Through a Concave Lens

REFRACTION THROUGH A CONCAVE LENS

Let us discuss the formation of images by a concave lens when the object is placed at two possible positions.

Object at Infinity

When an object is placed at infinity, a virtual image is formed at the focus. The size of the image is much smaller than that of the object (Figure 2.12).

Figure 2.12 Concave lens-Object at infinity

Object anywhere on the principal axis at a finite distance

When an object is placed at a finite distance from the lens, a virtual image is formed between optical center and focus of the concave lens. The size of the image is smaller than that of the object (Figure 2.13).

Figure 2.13 Concave lens-Object at a finite distance

But, as the distance between the object and the lens is decreased, the distance between the image and the lens also keeps decreasing. Further, the size of the image formed increases as the distance between the object and the lens is decreased. This is shown in (figure 2.14).

Figure 2.14 Concave lens- Variation in position and size of image with object distance

Understanding Image Formation by a Convex Lens: A Detailed Guide

Refraction Through a Convex Lens

10th Science : Chapter 2 : Optics : Refraction Through a Convex Lens

REFRACTION THROUGH A CONVEX LENS

Let us discuss the formation of images by a convex lens when the object is placed at various positions.

Object at infinity

When an object is placed at infinity, a real image is formed at the principal focus. The size of the image is much smaller than that of the object.

Figure 2.6: Object at infinity

Object placed beyond C (>2F)

When an object is placed behind the center of curvature(beyond C), a real and inverted image is formed between the center of curvature and the principal focus. The size of the image is the same as that of the object.

Figure 2.7: Object placed beyond C (>2F)

Object placed at C

When an object is placed at the center of curvature, a real and inverted image is formed at the other center of curvature. The size of the image is the same as that of the object.

Figure 2.8: Object placed at C

Object placed between F and C

When an object is placed in between the center of curvature and principal focus, a real and inverted image is formed behind the center of curvature. The size of the image is bigger than that of the object.

Figure 2.9: Object placed between F and C

Object placed at the principal focus F

When an object is placed at the focus, a real image is formed at infinity. The size of the image is much larger than that of the object.

Figure 2.10: Object placed at the principal focus F

Object placed between the principal focus F and optical centre O

When an object is placed in between principal focus and optical centre, a virtual image is formed. The size of the image is larger than that of the object.

Figure 2.11: Object placed between the principal focus F and optical centre O

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Image Formation by Convex and Concave Lenses: Rules of Refraction

Images Formed Due to Refraction Through a Convex and Concave Lens

When an object is placed in front of a lens, the light rays from the object fall on the lens. The position, size and nature of the image formed can be understood only if we know certain basic rules.

Rule 1:

When a ray of light strikes the convex or concave lens obliquely at its optical centre, it continues to follow its path without any deviation (Figure 2.3).

Rule 2:

When rays parallel to the principal axis strikes a convex or concave lens, the refracted rays are converged to (convex lens) or appear to diverge from (concave lens) the principal focus (Figure 2.4).

Rule 3:

When a ray passing through (convex lens) or directed towards (concave lens) the principal focus strikes a convex or concave lens, the refracted ray will be parallel to the principal axis (Figure 2.5).

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10th Science : Chapter 2 : Optics : Images Formed Due to Refraction Through a Convex and Concave Lens

Understanding Light Dispersion and Refraction of Composite Light

Refraction of a Composite Light-Dispersion of Light

REFRACTION OF A COMPOSITE LIGHT-DISPERSION OF LIGHT

We know that Sun is the fundamental and natural source of light. If a source of light produces a light of single colour, it is known as a monochromatic source. On the other hand, a composite source of light produces a white light which contains light of different colours. Sun light is a composite light which consists of light of various colours or wavelengths. Another example for a composite source is a mercury vapour lamp. What do you observe when a white light is refracted through a glass prism?

When a beam of white light or composite light is refracted through any transparent media such as glass or water, it is split into its component colours. This phenomenon is called as ‘dispersion of light’.

The band of colours is termed as spectrum. This spectrum consists of following colours: Violet, Indigo, Blue, Green, Yellow, Orange, and Red. These colours are represented by the acronym “VIBGYOR”. Why do we get the spectrum when white light is refracted by a transparent medium? This is because, different coloured lights are bent through different angles. That is the angle of refraction is different for different colours.

Angle of refraction is the smallest for red and the highest for violet. From Snell’s law, we know that the angle of refraction is determined in terms of the refractive index of the medium. Hence, the refractive index of the medium is different for different coloured lights. This indicates that the refractive index of a medium is dependent on the wavelength of the light.

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Course Context: 10th Science : Chapter 2 : Optics : Refraction of a Composite Light-Dispersion of Light

Properties of Light: A Comprehensive Guide for 10th Science Optics

Properties of Light

PROPERTIES OF LIGHT

Let us recall the properties of light and the important aspects on refraction of light.

1. Light is a form of energy.
2. Light always travels along a straight line.
3. Light does not need any medium for its propagation. It can even travel through vacuum.
4. The speed of light in vacuum or air is, c = 3 × 10⁸ ms^–1.
5. Since, light is in the form of waves, it is characterized by a wavelength (λ) and a frequency (ν), which are related by the following equation: c = ν λ (c - velocity of light).
6. Different coloured light has different wavelength and frequency.
7. Among the visible light, violet light has the lowest wavelength and red light has the highest wavelength.
8. When light is incident on the interface between two media, it is partly reflected and partly refracted.

Optics Introduction | Class 10 Science Chapter 2 Explained

Optics - Introduction

Light is a form of energy which travels in the form of waves. The path of light is called ray of light and group of these rays are called as beam of light. Any object which gives out light are termed as source of light. Some of the sources emit their own light and they are called as luminous objects. All the stars, including the Sun, are examples for luminous objects. We all know that we are able to see objects with the help of our eyes. But, we cannot see any object in a dark room. Can you explain why? If your answer is ‘we need light to see objects’, the next question is ‘if you make the light from a torch to fall on your eyes, will you be able to see the objects?’ Definitely, ‘NO’. We can see the objects only when the light is made to fall on the objects and the light reflected from the objects is viewed by our eyes. You would have studied about the reflection and refraction of light elaborately in your previous classes. In this chapter, we shall discuss about the scattering of light, images formed by convex and concave lenses, human eye and optical instruments such as telescopes and microscopes.

Tags: Introduction, 10th Science : Chapter 2 : Optics

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Topic: 10th Science : Chapter 2 : Optics : Optics | Introduction