Wave Optics

JEE Main level, single correct, medium

147 of 191

Question 147 of 191

A single slit of width bb is illuminated by a coherent monochromatic light of wavelength λ.\lambda . If the second and fourth minima in the diffraction pattern at a distance 1cm1cm from the slit are at 3cm3cm and 6cm6cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum)

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Answer: (D) 3.0cm3.0cm

For secondary minima dsin⁡⁡θ=nλd\sin ⁡\theta =n\lambda
sin⁡⁡θ=nλd\sin ⁡\theta =\dfrac{n\lambda }{d}

For second minima
n=2n=2
sin⁡⁡θ=2λd=tan⁡⁡θ1=x1D\sin ⁡\theta =\dfrac{2\lambda }{d}=\tan ⁡{\theta }_{1}=\dfrac{{x}_{1}}{D}
For fourth minima n=4n=4
sin⁡⁡θ2=4λd=x2D{\sin ⁡\theta }_{2}=\dfrac{4\lambda }{d}=\dfrac{{x}_{2}}{D}

Distance between both minima is x2−x1=4λd−2λd=2λd=6−3=3cm{x}_{2}-{x}_{1}=\dfrac{4\lambda }{d}-\dfrac{2\lambda }{d}=\dfrac{2\lambda }{d}= 6 -3 =3cm

Width of central maxima,=2λd=3cm=\dfrac{2\lambda }{d}=3\text{c}\text{m}

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