Question

A thin prism of angle $15$ made of glass of refractive index ${\mu _1} = 1.5$ is combined with another prism of glass of refractive index ${\mu _2} = 1.75$ . The combination of the prism produces dispersion without deviation. The angle of the second prism should be:
(A) $7$
(B) $10$
(C) $12$
(D) $5$

Answer 1

Hint : The above given two prisms are made up of two different glasses with different refractive indices and are combined. Total deviation is given as $zero$ . Use the given information in the formula for deviation as $deviation = \left( {\mu - 1} \right)\theta$ .

Complete Step By Step Answer:
From the above given data,
Let total deviation be, $\delta = 0$
Deviation of the first prism is: ${\delta _1} = ({\mu _1} - 1){\theta _1}$
Deviation of the second prism is: ${\delta _2} = ({\mu _2} - 1){\theta _2}$
But, according to the given data the total deviation is $zero$
$\delta = {\delta _1} + {\delta _2} = 0$
$\Rightarrow ({\mu _1} - 1){\theta _1} + ({\mu _2} - 1){\theta _2} = 0$ …..(substituting from above) $(1)$
Now we have given,
${\mu _1} = 1.5$ , ${\theta _1} = 15$ and ${\mu _2} = 1.75$ , ${\theta _2} = ?$
In equation $(1)$ put the above values and we get,
$\Rightarrow (1.5 - 1)15 + (1.75 - 1){\theta _2} = 0$
$\Rightarrow 0.5 \times 15 + 0.75 \times {\theta _2} = 0$
$\Rightarrow {\theta _2} = - \dfrac{{7.5}}{{0.75}} = - 10$
Here, we get the value of ${\theta _2}$ in negative value, it is because the negative sign shows that the prism two is oppositely combined with the prism one.
Thus the magnitude of ${\theta _2}$ is $10$
Thus the angle of the second prism is $10$
The correct answer is option B.

Note :
We know that the refractive index is the measure of the bending of a ray of light when passing from one medium into another. Angle of deviation is said to be the angle which is obtained from the difference between the angle of incidence and the angle of refraction created by the ray of light travelling from one medium to another that has a different refractive indices. Here, we have used the formula of deviation and reached our answer.