Question

$\large\sf\underline{Question:-}$

Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 × 10-6 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.

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Answer 1

ANSWER

• LET US CONSIDER POSITION OF FIRST MAXIMA FORMED BY WAVELENGTH 596 nm as y1
• LET US CONSIDER POSITION OF FIRST MAXIMA FORMED BY WAVELENGTH 590 as y2

$y{1} = (1 + \frac{1}{2} ) \frac{596 \: \: \: \: nm \: \: \: .d}{d}$

$y1 = (\frac{3}{2}) \frac{596 \: \: \: \: \: nm \: \: \: \: \: \: \: .d}{d}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ similarly$

$y2 = (1 + \frac{1}{2} ) \frac{590 \: \: \: \: \: nm \: \: \: \: \: \: \: .d}{d}$

$y2 = (\frac{3}{2} ) \frac{590 \: \: \: \: \: \: nm \: \: \: \: \: .d}{d}$

$calculating \: the \: required \: \\ seperation$

$y1 - y2$

$( \frac{3}{2} ) \frac{596 \: \: nm \: \: \: \: .d}{d} - (\frac{3}{2} ) \frac{590 \: \: nm \: \: .d}{d}$

$\frac{3.d}{2d} (596 \: \: nm - 590 \: \: nm$

$\frac{3.d}{2d} (6 \: \: nm)$

$\frac{18.d}{2d} nm$

$\frac{9.d}{d} nm$

$.d = 1.5m$

$d = 2 \times {10}^{ - 6} m$

$substituting \: values$

$\frac{9 \times 1.5 \times {10}^{ - 9} }{2 \times {10}^{ - 6} }$

$(:.1m = {10}^{ - 9} m)$

$\frac{9 \times 3 \times {10}^{6} }{2 \times 2 \times {10}^{ 9} }$

$\frac{27 \times {10}^{6 - 9} }{4}$

$6.75 \times {10}^{ - 3} m6.75mm$

$the \: seperation \: between \: the \\ \: position \: of \: first \: maxima \: \\ of \: the \: diffraction \: pattern \\ obtained \: in \: the \: two \: cases \: is \: \\ 6.75mm$

Answer 2

From 2010 to 2011, the sales of watch decreased by 80%. If the sales in 2012 was the same as in 2010, then what per cent did it increase from 2011 to 2012