Q85 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
A thin prism $P_1$ with an angle $6^{\circ}$ and made of glass of refractive index 1.54 is combined with another prism $\mathrm{P}_2$ made from glass of refractive index 1.72 to produce dispersion without average deviation. The angle of prism $\mathrm{P}_2$ is :
- A$6^{\circ}$
- B$1.3^{\circ}$
- C$7.8^{\circ}$
- Correct answerD$4.5^{\circ}$
Solution
Angle of deviation for first prism
$\delta_1=A_1\left(\mu_1-1\right)=6(1.54-1)$
Angle of deviation for second prism
$\begin{aligned}<br />& \delta_2=A_2\left(\mu_2-1\right) \\<br />& =A_2(1.72-1)<br />\end{aligned}$
For dispersion without deviation
$\begin{aligned}<br />& \delta_1=\delta_2 \\<br />& \Rightarrow 6^{\circ}(1.54-1)=\mathrm{A}_2(1.72-1) \\<br />& \Rightarrow \mathrm{A}_2=\frac{6^{\circ} \times 0.54}{0.72}=\frac{18^{\circ}}{4}=4.5^{\circ}<br />\end{aligned}$
$\delta_1=A_1\left(\mu_1-1\right)=6(1.54-1)$
Angle of deviation for second prism
$\begin{aligned}<br />& \delta_2=A_2\left(\mu_2-1\right) \\<br />& =A_2(1.72-1)<br />\end{aligned}$
For dispersion without deviation
$\begin{aligned}<br />& \delta_1=\delta_2 \\<br />& \Rightarrow 6^{\circ}(1.54-1)=\mathrm{A}_2(1.72-1) \\<br />& \Rightarrow \mathrm{A}_2=\frac{6^{\circ} \times 0.54}{0.72}=\frac{18^{\circ}}{4}=4.5^{\circ}<br />\end{aligned}$
Hence, the correct answer is option: 'D'.