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.