# 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 - Physics

Numerical

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 centre of the screen. Find the width of the slit.

#### Solution

Wavelength of light beam, λ = 500 nm = 500 × 10−9 m

Distance of the screen from the slit, D = 1 m

For first minima, n = 1

Distance between the slits = d

Distance of the first minimum from the centre of the screen can be obtained as:

x = 2.5 mm = 2.5 × 10−3 m

It is related to the order of minima as:

"n"lambda = "x" "d"/"D"

"d" = ("n"lambda"D")/"x"

= (1 xx 500 xx 10^(-9) xx 1)/(2.5 xx10^(-3))

= 2 × 10−4 m

= 0.2 mm

Therefore, the width of the slits is 0.2 mm.

Concept: Diffraction - Seeing the Single Slit Diffraction Pattern
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#### APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 10 Wave Optics
Q 19 | Page 385
NCERT Physics Part 1 and 2 Class 12
Chapter 10 Wave Optics
Exercise | Q 10.19 | Page 385