This question is about refractive index. Light from a laser is directed at a semi - circular glass block. The light passes undeviated through the block and on to a screen, forming a spot as shown. The semi - circular block is rotated about the point P. The spot of light on the screen is seen to move downwards. When the spot reaches point B, it disappears.

Engineering

Physics

Curved Surface Refraction

Question

This question is about refractive index. Light from a laser is directed at a semi - circular glass block. The light passes undeviated through the block and on to a screen, forming a spot as shown.

The semi - circular block is rotated about the point P. The spot of light on the screen is seen to move downwards. When the spot reaches point B, it disappears.

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Linked Question 1

In a particular experiment, the distance PA is 120 cm and distance AB is 90 cm. Calculate the refractive index of the glass of the block

A

$\frac{5}{3}$

B

None of these

C

$\frac{4}{3}$

D

$\frac{5}{4}$

$PB = \sqrt{{PA}^{2} + {AB}^{2}} = 150 cm$

$θ = 90 ° - i ∠ 90 °$

independent of radius.

disapearing relates to angle.

r = i + θ = 90°

$μ = \frac{\sin 90 °}{sini} = \frac{1}{cosθ}$

$μ = \frac{PB}{PA} = \frac{150}{120} = \frac{5}{4}$ .

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Linked Question 2

Which of the following diagram shows the position of the semi - circular block when the spot is at point B.

A

B

None of these

C

D

$PB = \sqrt{{PA}^{2} + {AB}^{2}} = 150 cm$

$θ = 90 ° - i ∠ 90 °$

independent of radius.

disapearing relates to angle.

r = i + θ = 90°

$μ = \frac{\sin 90 °}{sini} = \frac{1}{cosθ}$

$μ = \frac{PB}{PA} = \frac{150}{120} = \frac{5}{4}$ .

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