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Question

Physics Question on Wave optics

A screen is placed 50 cm50\, cm from a single slit which is illuminated with light of wavelength 6000?6000 ?. If the distance between the first and third minima in the diffraction pattern is 3.0 mm3.0 \,mm. The width of the slit is

A

1×10−4 m1 \times 10^{-4} \,m

B

2×10−4 m2 \times 10^{-4} \,m

C

0.5×10−4 m0.5 \times 10^{-4} \,m

D

4×10−4 m4 \times 10^{-4} \,m

Answer

2×10−4 m2 \times 10^{-4} \,m

Explanation

Solution

The position of nthn^{th} minima in the diffraction pattern is xn=nDλdx_{n} = \frac{nD\lambda}{d} ∴x3−x1=(3−1)Dλd\therefore x_{3} -x_{1} = \left(3-1\right) \frac{D\lambda}{d} =2Dλd= \frac{2D\lambda}{d} or d=2Dλx3−x1d =\frac{ 2D\lambda}{x_{3} -x_{1} } =2×0.50×6000×10−103×10−3 =\frac{ 2\times 0.50 \times 6000 \times 10^{-10}}{3\times 10^{-3}} =2×10−4m= 2 \times 10^{-4} m