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# Monochromatic light of wavelength $5000\;A^{\circ}$ is used in Y.D.S.E with slit- width $d= 1 mm$distance between screen and slits $D=1\;m$ .Then the distance of 5th minima from central maxima on screen is :-

$(a)\;1.25\;mm \\ (b)\;1\;m \\ (c)\;2\;m \\ (d)\;2.25 \;mm$

For minima
$y= (2n-1) \large\frac{\lambda D}{d}$
$n=5$
$\large\frac{\lambda D}{d}=\frac{5000 \times 10^{-10} \times 1}{1 \times 10^{-3}}$$=0.5 mm$
$y= 2.25 mm$
Hence d is the correct answer.

edited Jul 22, 2014