Question

A thin prism $P_1$ with angle $6^{\circ }$ and made from glass of refractive index $1.54$ is combined with another thin prism $P_2$ of refractive index $1.72$ to produce dispersion without deviation. The angle of prism $P_2$ will be

• A. $5^{\circ }{24}^{\prime }$
• B. $4^{\circ }{30}^{\prime }$
• C. $6^{\circ }$
• D. $8^{\circ }$

Answer 1

The correct option is B $4^{\circ }{30}^{\prime }$

Let $\delta$ and ${\delta }^{\prime }$ is deviation produced by prism $P_1$ and $P_2$ respectively.

To produce dispersion without deviation,

$\delta +{\delta }^{\prime }=0$

$\Rightarrow \delta =-{\delta }^{\prime }$

or, $A(\mu -1)=-A^{\prime }({\mu }^{\prime }-1)$

or, $\frac{A^{\prime }}{A}=\frac{-(\mu -1)}{({\mu }^{\prime }-1)}$

For prism $P_1,A=6^{\circ }$ and $\mu =1.54$.

Also for prism $P_2,{\mu }^{\prime }=1.72$

$\Rightarrow \frac{A^{\prime }}{6}=\frac{-(1.54-1)}{1.72-1}$

or $\frac{A^{\prime }}{6}=\frac{-0.54}{0.72}$

or, $\frac{A^{\prime }}{6}=-\frac{3}{4}$

$\therefore A^{\prime }=\frac{-18}{4}=-{4.5}^{\circ }$

$\Rightarrow A^{\prime }=-4^{\circ }{30}^{\prime }$

The negative sign indicates that prism $P_2$ is inverted with respect to prism $P_1$.

Why this question? Tip: To obtain dispersion without deviation, combination of two prisms inverted with respect to each other is required.