Question 39: Waves Optics

Question

In the young's double slit experiment the fringe width of the interference pattern is found to be $$3.2 	imes 10^{-4} \mathrm{~m}$$, when the light of wave length $$6400^{\circ} \mathrm{A}$$ is used. What will be change in fringe width if the light is replaced with a light of wave length $$4800^{\circ} \mathrm{A}$$

Accepted Answer

To find the linear width of the central maximum in a diffraction pattern, we can use the formula related to the diffraction due to a single slit. The angular width of the central maximum is given by $$	heta = 2 \lambda / b$$, where $$\lambda$$ is the wavelength of the light (in meters) and $$b$$ is the width of the slit (in meters). The linear width of the central maximum on the screen (which we'll denote as $$W$$) is then given by $$W = 2L	an(	heta/2) \approx 2L (	heta/2)$$ for small angles, where $$L$$ is the distance from the slit to the screen.
Given:
The wavelength of light, $$\lambda = 6400^{\circ} A = 6400 	imes 10^{-10} m = 640 	imes 10^{-9} m$$ (since $$1 \mathring{A} = 10^{-10} m$$).
The width of the slit, $$b = 2 mm = 2 	imes 10^{-3} m$$.
The distance from the slit to the screen, $$L = 2 m$$.
First, calculate the angular width of the central maximum:
$$	heta = 2 	imes \frac{\lambda}{b} = 2 	imes \frac{640 	imes 10^{-9} m}{2 	imes 10^{-3} m} = 640 	imes 10^{-6} rad.$$
For small angles, where $$	heta$$ is in radians, the approximation $$	an(	heta/2) \approx 	heta/2$$ can be used. Thus, the linear width of the central maximum on the screen is:
$$W \approx 2L 	imes \frac{	heta}{2} = 2 	imes 2 m 	imes 320 	imes 10^{-6} rad = 1280 	imes 10^{-6} m = 1.28 	imes 10^{-3} m = 1.28 mm.$$
Therefore, the correct answer is
Option B
: 1.28 mm.

Answer

$$2.4 	imes 10^{-4} \mathrm{~m}$$

Answer

$$1.6 	imes 10^{-4} \mathrm{~m}$$

Answer

$$0.8 	imes 10^{-4} \mathrm{~m}$$

Answer

$$5.6 	imes 10^{-4} \mathrm{~m}$$

Waves Optics

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