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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics1 Mark
MCQ
In Young's double slit experiment, $8^{\text {th }}$ maximum with wavelength ' $\lambda_1$ ' is at distance ' $d_1$ ' from the central maximum and $6^{\text {th }}$ maximum with wavelength ' $\lambda_2$ ' is at a distance ' $d_2$. Then $\frac{d_2}{d_1}$ is
  • A
    $\frac{3 \lambda_1}{4 \lambda_2}$
  • ✓
    $\frac{3 \lambda_2}{4 \lambda_1}$
  • C
    $\frac{4 \lambda_1}{3 \lambda_2}$
  • D
    $\frac{4 \lambda_2}{3 \lambda_1}$
✓

Answer

Correct option: B.
$\frac{3 \lambda_2}{4 \lambda_1}$
(b) : Position of $n^{\text {th }}$ maxima from central maxima is
$
x_n=\frac{n \lambda \cdot D}{d} \Rightarrow d=\frac{n \lambda D}{x_n}
$
As, same slit is used in both cases.
$
\therefore \frac{8 D \lambda_1}{d_1}=\frac{6 D \lambda_2}{d_2} ; \frac{d_2}{d_1}=\frac{6}{8} \frac{\lambda_2}{\lambda_1}=\frac{3}{4} \frac{\lambda_2}{\lambda_1}
$

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