CBSE 2022 · Region 4 · Set 1 · Q10 · 3 marks
A ray of light is incident on a prism at an angle of $\displaystyle 45$° and passes symmetrically as shown in the figure. Calculate:
(a)the angle of minimum deviation,(b)the refractive index of the material of the prism, and(c)the angle of refraction at the point P.
A ray of light is incident on a prism at an angle of $\displaystyle 45$° and passes symmetrically as shown in the figure. Calculate:
(a)
the angle of minimum deviation,
(b)
the refractive index of the material of the prism, and
(c)
the angle of refraction at the point P.
Marking-scheme solution
(a) As the ray passes symmetrically
$$\angle i=\angle e \text { and } \delta=\delta_{\mathrm{m}}
\begin{aligned}
\therefore \quad i+e & =A+\delta \\
2 i & =A+\delta_{m} \\
\delta_{m} & =2 i-A=90^{\circ}-60^{\circ} \\
& =30^{\circ}
\end{aligned}
(b) \begin{aligned}
\mu & =\frac{\sin \dfrac{\left(A+\delta_{m}\right)}{2}}{\sin A / 2} \\
& =\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\sqrt{2}
\end{aligned}
$$
(c)
$$\begin{aligned}
& \text { As } \boldsymbol{\delta}=\boldsymbol{\delta}_{\boldsymbol{m}}
& \mathrm{r}_{$\displaystyle 1$}=\mathrm{r}_{$\displaystyle 2$} \text { (Altern }
& \text { Hence, } \mathrm{r}_{$\displaystyle 1$} \text { (Angle o }
& $\displaystyle 1$ \times \sin $\displaystyle 45$=\mu \operatorname{sinr}_{$\displaystyle 1$}
& $\displaystyle 1$ \times \frac{1}{\sqrt{2}}=\sqrt{2} \mathrm{sin}_{$\displaystyle 1$}
& \therefore \angle r_{$\displaystyle 1$}=$\displaystyle 30$^{\circ}
\end{aligned}
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.