Ncert Exercises & Solutions

10.19 A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the center of the screen. Find the width of the slit.

Hint: $n\lambda =x\frac{d}{D}$

Step 1: Find the width of the slit.

Wavelength of light beam, λ = 500 nm = 500 × $10^{- 9}$ m
The distance of the screen from the slit, D = 1 m
For the first minima, n = 1

The distance of the first minimum from the center of the screen,
$x=2.5~mm.=2.5\times 10^{-3}~m.$
$Now,$

$\begin{array}{l} n λ = x \frac{d}{D} \\ d = \frac{n λ D}{x} \\ = \frac{1 \times 500 \times 10^{- 9} \times 1}{2.5 \times 10^{- 3}} = 2 \times 10^{- 4} m = 0.2 m m. \end{array}$

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