(a) With the help of labelled ray diagram show the path followed by a narrow beam of monochromatic light when it passes through a glass prism.

(b) What would happen if this beam is replaced by a narrow beam of white light ?

OR

(a) A person is suffering from both myopia and hypermetropia.

(i) What kind of lenses can correct this defect ?

(ii) How are these lenses prepared ?

(b) A person needs a lens of power + 3D for correcting his near vision and - 3D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.

Part (a)

When a narrow beam of monochromatic light passes through a glass prism, it bends towards the normal at the first refracting surface (air to glass) and away from the normal at the second refracting surface (glass to air). Since it is monochromatic (single wavelength), it does not split into multiple colours but simply deviates from its original path.

Part (b)

If the monochromatic beam is replaced by a narrow beam of white light, dispersion occurs. The white light splits into its seven constituent colours (VIBGYOR: Violet, Indigo, Blue, Green, Yellow, Orange, Red) forming a spectrum. This happens because different colours of light have different wavelengths and bend through different angles with respect to the incident ray as they pass through the prism (red bends the least, violet bends the most).

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OR

Part (a)

(i) A person suffering from both myopia and hypermetropia can correct this defect using bi-focal lenses.

(ii) These lenses are prepared such that the upper portion consists of a concave lens (to facilitate distant vision) and the lower portion consists of a convex lens (to facilitate near vision).

Part (b)

Given:

• Power of lens for near vision, $P_1 = +3\text{ D}$
• Power of lens for distant vision, $P_2 = -3\text{ D}$

1. Focal length for near vision ($f_1$):

$$f_1 = \frac{1}{P_1} = \frac{1}{+3}\text{ m} \approx +0.333\text{ m} = +33.3\text{ cm}$$

Since the focal length is positive, it is a converging (convex) lens.

2. Focal length for distant vision ($f_2$):

$$f_2 = \frac{1}{P_2} = \frac{1}{-3}\text{ m} \approx -0.333\text{ m} = -33.3\text{ cm}$$

Since the focal length is negative, it is a diverging (concave) lens.