Question

If a thin prism of glass is dipped into water then minimum deviation (with respect to air) of light produced by prism will be left \left( _{a}{{\mu }_{g}}=\frac{3}{2}\text{ and}{{\,}_{a}}{{\mu }_{w}}\,=\,\frac{4}{3} \right)
[UPSEAT 1999]
A) \frac{1}{2} B) \frac{1}{4} C) 2 D) \frac{1}{5}

Answer 1

Formula of deviation is given by
δ = (μ - 1)A
Where δ denotes the deviation , A is the angle of prism and μ is the refractive index of prism with respect to medium .

Here apply , $\bold{\frac{\delta_a}{\delta_w}=\frac{\frac{\mu_g}{\mu_a}-1}{\frac{\mu_g}{\mu_w}-1}}$
Here μa denotes refractive index of air e.g., μa = 1
μg denotes refractive index of glass e.g., μg = 3/2
μw denotes refractive index of water e.g,. μw = 4/3

Now, δa/δw = (3/2 - 1)/(3/2/4/3 - 1)
= (1 /2)/(9/8 - 1) = 4

Hence, minimum deviation in water = 1/4th of minimum deviation in air

Answer 2

Answer:

Explanation:

For thin prism, angle of minimum deviation δm=( µ-1) A

In air δm=( µg-1) A

In air δm=[(3/2)-1]A --eq(1)

In water δm=( µgw-1) A

Here µgw is refractive index of glass with respect to water

µgw=µg / µw

µgw=(3/2) / (3/4)=2

In water δm=(2-1)A=A --eq(2)

taking ratio of eq(1) and eq(2) we get answer 1/2