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,
]

Text Solution
Verified by ExpertsA

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

Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems