Wave Optics

JEE Main level, single correct, medium

109 of 191

Question 109 of 191

In Young’s double-slit experiment, the distance between the two identical slits is6.16.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is :

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Answer: (D) 1212

d=d = separation between 2 slits

a=a = slit width

βy={\beta }_{y}= fringe width in YDSE & βD={\beta }_{D}= width of central maxima

βy=λDd=λD6.1aβD=2λDa{\beta }_{y}=\dfrac{\lambda D}{d}=\dfrac{\lambda D}{6.1a} {\beta }_{D}=\dfrac{2\lambda D}{a}

βD=2×6.1βy{\beta }_{D}=2\times 6.1 {\beta }_{y}

=12.2βy=12.2 {\beta }_{y}

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