1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The angular width of the central maximum in a single slit diffraction pattern is 60∘. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young's Fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm. What is slit separation distance? (i.e. distance between the centre of each slit)

A
25.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
25.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
25
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## In single slit diffraction, The angular width of central maximum is given by 2θ Where, θ−angular position of first minimum Given that, 2θ=60∘ ⇒θ=30∘ The condition or first minimum is given by, asinθ=λ ⇒1×10−6sin30∘=λ ⇒λ=0.5×10−6m=5×10−7m In YDSE, Fringe width, β=λDd Where d = distance between the slits ⇒1×10−2=5×10−7×50×10−2d ⇒d=25×10−6 m=25 μm Why this question?Asked in JEE mains 2018.

Suggest Corrections
11
Join BYJU'S Learning Program
Related Videos
Diffraction I
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program