White coherent light (400 nm-700 nm) is sent through the slits of a Young’s double slit experiment (as shown in the figure). The separation between the slits is 1 mm and the screen is 100 cm away from the slits. There is a hole in the screen at a point 1.5 mm away (along the width of the fringes) from the central line.
Text Solution
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428 nm, 600 nm, 500 nm
Sol. As λ << d, ⇒ β = 
can be used
For minima
y = 
β .= 
⇒ λ = 
n = 1, λ = 
× 1.5 × 10 –6 = 1000 nm
n = 2, λ = 
× 1.5 × 10 –6 = 600 nm
n = 3, λ = 
× 1.5 × 10 –6 = 428 nm
Putting integral values of ‘n’
n = 1 ; λ = 1000 nm
n = 2; λ = 600 nm.
n = 3; λ = 428 nm
So only λ = 428 nm and λ = 600 nm, will have minima at the hole. Hence they will be absent in the light coming out.
1.5 mm = n β .
1.5 mm = 
⇒ λ = 
for n = 1. λ = 1500 nm
for n = 2. λ = 750 nm.
for n = 3. λ = 500 nm.
Hence only λ = 500 nm will have maximum intensity.
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