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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Wave Optics

JIPMER 2016 · Physics · Wave Optics

Angular width of central maximum in the Fraunhoffer diffraction...

Angular width of central maximum in the Fraunhoffer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000A˚6000 Å. When the slit is illuminated by light of another wavelength, then the angular width decreases by 30%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

Options

  1. A1.25
  2. B1.42
  3. C1.67
  4. D1.5

Correct answer

B. 1.42

Step-by-step solution

Given, λ1=6000A˚\lambda_1=6000 Å, λ2=?Angularwidth\lambda_2=?Angular width of central maximum =2λa\frac{2 \lambda}{a} \theta_1=2λ1a....(i)\frac{2 \lambda_1}{a}....(i) and \theta_2=2λ2a\frac{2 \lambda_2}{a} \because \quad \theta_2=\theta_1 \times 0.7 \therefore \quad \theta_1 \times 0.7=2λ2a.....(ii)From\frac{2 \lambda_2}{a}.....(ii) FromEqs. (i) and (ii), we get10.7=λ1λ2⇒λ2=λ1×0.7=6000×0.7=4200A˚\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}whenimmersedinliquid,when immersed in liquid,λ2=λ1μ∵θ2=θ1×0.7\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}×0.7\times 0.7 \quad\left[\because \lambda_2=\frac{\lambda_1}{\mu}\right] ⇒2λ1/μa=2λ1a×0.7⇒1μ=0.7⇒μ=1.42\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}$
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