Question

Draw the path of ray of light when it enters one of the faces of a glass slab at an angle of nearly 45 label on it (i) angle of refraction (ii)angle of emergence (iii) lateral displacement

Answer 1

Given that,

The path of ray of light when it enters one of the faces of a glass slab at an angle of nearly 45 label on it.

Suppose, the thickness of slab is 2 mm.

We know that,

Refractive index of air and glass are 1 and 1.5

The angle of incidence is equal to the angle of emergence.

(I), We need to calculate the angle of refraction

Using snell's law

$n_{1}\sin\theta_{1}=n_{2}\sin\theta_{r}$

$\sin\theta_{r}=\dfrac{n_{1}\sin\theta_{i}}{n_{2}}$

Put the value into the formula

$\sin\theta_{r}=\dfrac{1\times\sin(45)}{1.5}$

$\theta_{r}=\sin^{-1}(\dfrac{1\times\sin(45)}{1.5})$

$\theta_{r}=28.13^{\circ}$

(II). We need to calculate the angle of emergence

Using formula of emergence

$i=e$

Put the value into the formula

$e=45^{\circ}$

(III). We need to calculate the lateral displacement

Using formula of lateral displacement

$\text{lateral displacement}=\dfrac{t\sin(i-r)}{\cos r}$

Where, t = thickness of slab

Put the value into the formula

$\text{lateral displacement}=\dfrac{2\times10^{-3}\sin(45-28.13)}{\cos 28.13}$

$\text{lateral displacement}=0.66\ mm$

Hence, (I). The angle of refraction is 28.13°

(II). The angle of emergence is 45°

(III). The lateral displacement is 0.66 mm.

Answer 2

Answer:

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Explanation:

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