Home Physics Wave Optics NTA Abhiyas Question A screen is at a distance D = 80 cm from a d…
Physics Wave Optics NTA Abhiyas Question MCQ (Single Correct)

A screen is at a distance D = 80 cm from a diaphragm having two narrow slits S 1 and S 2 which are d = 2 mm apart. The slit Si is covered by a transparent sheet of thickness t 1 = 2.5 m and the slit 62 by another sheet of thickness t 2 = 1.25 m as shown in the figure. Both sheets are made of the same material having a refractive index m = 1.40. Water is filled in space between the diaphragm and the screen. A monochromatic light beam of wavelength = 5000 A is incident normally on the diaphragm. Assuming the intensity of the beam to be uniform and slits of equal width, calculate the ratio of intensity at C to the maximum intensity of interference pattern obtained on the screen, where C is the foot of the perpendicular bisector of S 1 S 2 . [Refractive index of water, ]

A
B
C
D

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Text Solution

Verified by Experts
The correct answer is:
A

Optical path = (Refractive Index) x (Geometrical Path Length)

Path Difference at point C on the screen

Phase difference

at any point on the screen

Resultant intensity I C = I + I +

I C = 3I

& I max = 1 + 1 + 21 when = 0° cos = +1

I max = 41

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