Q120 / 180 · JIPMER 2016 · Physics

Angular width of central maximum in the Fraunhoffer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength $6000 Å$. When the slit is illuminated by light of another wavelength, then the angular width decreases by $30 \%$. The same decrease in angular width of central maximum is obtained when the original apparatus is immersed in a liquid. The refractive index of the liquid will be

  • A
    1.25
  • Correct answer
    B
    1.42
  • C
    1.67
  • D
    1.5

Solution

Given, $\lambda_1=6000 Å$, $\lambda_2=?$ Angular width of central maximum $=\frac{2 \lambda}{a}$ $\theta_1=\frac{2 \lambda_1}{a}$....(i) and $\theta_2=\frac{2 \lambda_2}{a}$ $\because \quad \theta_2=\theta_1 \times 0.7$ $\therefore \quad \theta_1 \times 0.7=\frac{2 \lambda_2}{a}$.....(ii) From Eqs. (i) and (ii), we get $\begin{aligned} \frac{1}{0.7} & =\frac{\lambda_1}{\lambda_2} \\ \Rightarrow \quad \lambda_2 & =\lambda_1 \times 0.7=6000 \times 0.7 \\ & =4200 Å\end{aligned}$ when immersed in liquid, $\begin{aligned} \lambda_2 & =\frac{\lambda_1}{\mu} \\ \because \quad \theta_2 & =\theta_1 \times 0.7\end{aligned}$ $\Rightarrow \quad \frac{2 \lambda_2}{a}=\frac{2 \lambda_1}{a} \times 0.7 \quad\left[\because \lambda_2=\frac{\lambda_1}{\mu}\right]$ $\begin{aligned} \Rightarrow & \frac{2 \lambda_1 / \mu}{a} & =\frac{2 \lambda_1}{a} \times 0.7 \\ \Rightarrow & \frac{1}{\mu} & =0.7 \Rightarrow \mu=1.42\end{aligned}$

Hence, the correct answer is option: 'B'.

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