Question

A prism of angle $3{}^\circ$ is made of glass having refractive index $1.64$. Two thin prisms of glass having refractive index $1.48$ are intended to be coupled with the former prism to yield a combination without an average deviation. Which one of the following angles cannot correspond to the two prisms?
$A)2{}^\circ$ and $2{}^\circ$
$B)1.5{}^\circ$ and $2.5{}^\circ$
$C)6{}^\circ$ and $2{}^\circ$
$D)5.5{}^\circ$ and $3.5{}^\circ$

Answer 1

The deviation caused due to the first prism is calculated. Similarly, the deviations caused by the second and the third prism are determined. Since the average deviation caused by the first prism, coupled with the combination of second and third prism is given to be zero, the net deviation caused due to the combination of the second prism and the third prism can be equated to the deviation caused by the first prism. Finally, the sum of angles of prism of the second prism and the third prism is tallied with the sum of angles provided in the options.

Formula used:
$\delta =A(\mu -1)$

Complete answer:
We know that when light falls on a prism, the light gets deviated after passing through the prism. A total of two refractions happen inside the glass prism to bring out a deviation in the incident light. The deviation caused by a prism is mathematically given by
$\delta =A(\mu -1)$
where
$\delta$ is the deviation in incident light caused due to a prism
$A$ is the angle of the prism
$\mu$ is the refractive index of the material of the prism
Let this be equation 1.
Coming to our question, we are told that a prism is coupled with a combination of other two prisms of equal refractive index. We are required to find the angles of these two prisms. We are provided with the angle of the first prism and the refractive index of the material of the first prism as $3{}^\circ$ and $1.64$, respectively. We are also given that the equal refractive index of the second prism and the third prism is $1.48$.
Using equation 1, if the deviation caused by the first prism is denoted as ${{\delta }_{1}}$, it is given by
${{\delta }_{1}}={{A}_{1}}({{\mu }_{1}}-1)$
where
${{\delta }_{1}}$ is the deviation caused by the first prism
We have assumed that ${{A}_{1}}$ is the angle of the first prism
We have assumed that ${{\mu }_{1}}$ is the refractive index of the material of the first prism
Clearly, from the values provided in the question, the above expression turns out to be
${{\delta }_{1}}={{A}_{1}}({{\mu }_{1}}-1)=3(1.64-1)=1.92$
Let this be equation 2.
Similarly, using equation 1, the deviations caused by second prism and third prism are given by
$\begin{aligned} & {{\delta }_{2}}={{A}_{2}}({{\mu }_{2}}-1)={{A}_{2}}(1.48-1)=0.48{{A}_{2}} \\\ & {{\delta }_{3}}={{A}_{3}}({{\mu }_{3}}-1)={{A}_{3}}(1.48-1)=0.48{{A}_{3}} \\\ \end{aligned}$
where
${{\delta }_{2}}$ is the deviation caused due to the second prism
${{\delta }_{3}}$ is the deviation caused due to the third prism
${{A}_{2}}$ is the angle of the second prism
${{A}_{3}}$ is the angle of the third prism
Let the above equations be equation 3 and equation 4, respectively.
Now, we are required to determine the net deviation caused due to the combination of the second prism and the third prism. If the net deviation of the combination of the second prism and the third prism is denoted as ${{\delta }_{net}}$, it is given by
${{\delta }_{net}}={{\delta }_{2}}+{{\delta }_{3}}=0.48({{A}_{2}}+{{A}_{3}})$
Let this be equation 5.
Now, we are told that the average deviation caused when the first prism is coupled with the combination of the second prism and the third prism is equal to zero. If the average deviation of the first prism coupled with the combination of the second prism and the third prism is represented by ${{\delta }_{avg}}$, it is given that
${{\delta }_{avg}}={{\delta }_{1}}-{{\delta }_{net}}=0\Rightarrow {{\delta }_{1}}={{\delta }_{net}}$
Let this be equation 6.
Clearly, equation 6 tells that the deviation caused by the first prism is equal to the deviation caused by the combination of the second prism and the third prism. Substituting equation 2 and equation 5 in equation 6, we have
${{\delta }_{1}}={{\delta }_{net}}\Rightarrow 1.92=0.48({{A}_{2}}+{{A}_{3}})\Rightarrow {{A}_{2}}+{{A}_{3}}=\dfrac{1.92}{0.48}=4$
Let this be equation 7.
Therefore, the sum of angle of the second prism and the angle of the third prism is equal to $4$.
Now, in the options provided, we are given with the angles of both the second prism and the third prism. But what we got using our calculations is the sum of both these angles. So, we need to cross check each option to see if the sum of angles of both the prisms is equal to $4$ or not. Clearly, option $D)5.5{}^\circ$ and $3.5{}^\circ$ does not satisfy the condition.

Therefore, the correct option to be marked is $D$.

Note: Students need not get confused with option $C$. They should understand that the second prism and the third prism can also be placed in an opposite manner, to bring out the same deviation. When this is the case, the difference in the angle of the second prism and the angle of the third prism turns out to be $4$ itself. Therefore, option $C$ also satisfies the given condition.