Ncert Exercises & Solutions

10.4 In Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Hint: For constructive interference

$u=n\lambda\frac{D}{d}$

Step 1: Find the wavelength of light used in the experiment.
Distance between the slits and the screen, D = 1.4 m
and the distance between the slits, d = 0.28 mm = 0.28 x 10^-3 m
Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 × 10^-2 m $\begin{array}{l} \end{array}$

$As~u=n\lambda\frac{D}{d}$
$\Rightarrow\lambda =\frac{ud}{nD} =\frac{1.2\times 10^{-2}\times 0.28\times 10^{-3}}{4\times 1.4}=6\times10^{-7}~m=600~nm$

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