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In a Young’s double slit experiment, slits are illuminated by a monochromatic source of wavelength $6000 \overset{0}{A}$ and fringes are obtained. If screen is moved by a distance of 5 cm towards slits, change in fringe width is $3 \times 10^{- 5}$ m , then what will be the separation (in mm) between the slits?

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a

2

b

1

c

3

d

4

answer is A.

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Detailed Solution

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$L e t s a y o l d f r i n g e w i d t h i s β = \frac{λ D}{d} n e w f r i n g e w i d t h i s β' = \frac{λ D'}{d} D i f f e r e n c e b e t w e e n f r i n g e w i d t h s β - β' = \frac{λ (D - D')}{d} \Rightarrow d = \frac{λ (D - D')}{(β - β')} = \frac{6000 \times 10^{- 10} \times 5 \times 10^{- 2}}{3 \times 10^{- 5}} m = 10^{- 3} m = 1 m m$

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