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In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

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In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1m away. A monochromatic light of wave length 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

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(a) In Young's double slit experiment, two slits are 1 mm apart and the screen is placed 1 m away from the slits. Calculate the fringe width when light of wavelength 500 nm is used. (b) What should be the width of each slit in order to obtain 10 maxima of the double slits pattern within the central maximum of the single slit pattern?

Fringe width is given by $$\beta = \frac{\lambda d}{d}$$ $$<= \frac{500 \times 10^{-9} \times 1}{10^{-3}} = 0.5 mm = 0.5 \times 10^{-3} m = 5 \times 10^{-4} m$$ (b) $$\beta_0 = \frac{2\lambda d}{a} = 10 \beta$$ $$\rightarrow a = \frac{2 \times 500 \times 10^{-9} \times 1}{10 \times 5 \times 10^{-4}} = 2 \times 10^{-4} m$$.

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In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 mm is used. What will be the width of each slit for obtaining tenth maxima of double slit within the central maxima of single slit pattern ?

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Correct option is C

Distance between two slits d=1mm=10 -3 mm ; D=1m λ = 500 × 10 − 9 m  Width of each slit a = Width of central maxima in single slit pattern = 2 λD a ;  Condition is  10 λD d = 2 λD a a = d 5 D = 1 5 × 10 − 3 m ; a = 0 .2 mm

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In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic lightg of wavelength 500 nm is used, what will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern ?

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The correct Answer is: D

Angular width of the central maximum in a double slit experiment is x d = λ d d / d = λ d and the angular width of the central maximum in a single slit diffraction pattern is 2 λ a where 'a' is the width of the slit. it is given that 2 λ a = 10 λ d . ∴ a = 2 λ d 10 λ = .2 d = 0.2 × 1 = 0.2 m m, topper's solved these questions, similar questions.

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