Question

A coherent parallel beam of light of wavelength 5000 a is incident normally on the plane of slits. The slit s1 is covered with a slab of refractive index 1.5 and thickness has shown. The separation between the slit is and the separation between the plate of the slit is then the position of central maxima formed on the screen ends

Answer 1

The position of central maxima formed on the screen ends 2 cm below point "O".

Explanation:

Given data:

Wavelength of light = 5000 m

Refractive index "n" = 1.5

For central maxima dsinθ = (μs−1)t

1×10^−3 sinθ=(1.5−1) × 10^−5

1×10^−3 sinθ = 2×10^−5

sinθ = 1 / 50

tanθ =  1 / 50  ("for small 'θ')

y / D = 1 / 50  

Since y = D/ 50 = 100 / 5 cm

          = 2 cm

Thus the position of central maxima formed on the screen ends 2cm below point "O".

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