Light of wavelength 500 nm is incident on a slit o... moref width 0.1 ...
Beta(central Maxima)=2lambda D(distance of the screen)/d(distance between slits).
beta=2m.
wavelength= 500nm=5×10^-7m.
d=0.1mm=1×10^-4m.
2=2×5×10^-7 D/10^-4.
1=5×10^-3 D .
1/5×10^-3=D .
1000/5=D.
200m=D
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Light of wavelength 500 nm is incident on a slit o... moref width 0.1 ...
Given:
Wavelength of light, λ = 500 nm = 5 x 10^-7 m
Width of the slit, a = 0.1 mm = 10^-4 m
Width of central bright line, y = 2 m
To find:
Distance of the screen, D
Formula used:
y = (λD) / a
Calculation:
y = 2 m
λ = 5 x 10^-7 m
a = 10^-4 m
Using the formula,
y = (λD) / a
2 = (5 x 10^-7 x D) / 10^-4
2 x 10^-4 = 5 x 10^-7 x D
D = (2 x 10^-4) / (5 x 10^-7)
D = 400000 / 5
D = 80000 m = 200 m
Therefore, the distance of the screen is 200 m.
Answer:
Option (B) 200 m is the correct answer.
Light of wavelength 500 nm is incident on a slit o... moref width 0.1 ...
Beta(central Maxima)=2lambda D(distance of the screen)/d(distance between slits).beta=2m.wavelength= 500nm=5×10^-7m.d=0.1mm=1×10^-4m.2=2×5×10^-7 D/10^-4.1=5×10^-3 D .1/5×10^-3=D .1000/5=D.200m=DI think this is the answer.....