1985-5 (Physical Optics, Geometric Optics) Light of wavelength 5.0 × 10 -7 m in

Question

1985-5 (Physical Optics, Geometric Optics)

Light of wavelength 5.0 × 10-7 m in air is incident normally (perpendicularly) on a double slit. The distance between the slits is 4.0 × 10-4 m, and the width of each slit is negligible. Bright and dark fringes are observed on a screen 2.0 m away from the slits.

a. Calculate the distance between two adjacent bright fringes on the screen. The entire double-slit apparatus, including the slits and the screen, is submerged in water, which has an index of refraction 1.3.

b. Determine each of the following for this light in water.

i. The wavelength

ii. The frequency

c. State whether the distance between the fringes on the screen increases, decreases, or remains the same. Justify your answer.

Please explain all Answers

Explanation / Answer

(a) Use the expression -

x = (/d) * D

given -
d = slit width = 4.0*10^-4 m
D = Distance to screen = 2 m
x = [(5.0x10^-7) / (4.0*10^-4)]*2 = 2.5 x 10^-3 m 2.5 mm

(b) ' = 1/1.3 = (1/1.3)*(5.0x10^-7) = 3.84 x10^-7 m

c = f *
f = c/ = (3.0x10^8) / (5.0x10^-7) = 6.0 x 10^14 Hz

(c)  Decreases to 1/1.3 = 77%

Because, is now smaller.

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