(B) The distance between the two first-order minima on either side of the central maximum is equal to the width of the central maximum.The formula for

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(B) The distance between the two first-order minima on either side of the central maximum is equal to the width of the central maximum.The formula for the width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2 \lambda D}{a}$.Given values are:Slit width $a = 2 \,mm = 2 \times 10^{-3} \,m$Distance to screen $D = 2 \,m$Wavelength $\lambda = 6330 \mathring{A} = 6330 \times 10^{-10} \,m$Substituting these values into the formula:$w = \frac{2 \times 6330 \times 10^{-10} \times 2}{2 \times 10^{-3}}$$w = 2 \times 6330 \times 10^{-7} \,m$$w = 12660 \times 10^{-7} \,m = 1.266 \times 10^{-3} \,m$$w \approx 1.27 \,mm$.

Class 12 Physics Wave Optics Single Slit Diffraction of Light

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Through a narrow slit of width $2 \,mm$, a diffraction pattern is formed on a screen kept at a distance $2 \,m$ from the slit. The wavelength of the light used is $6330 \mathring{A}$ and it falls normal to the slit and screen. Then, the distance between the two minima on either side of the central maximum is (in $\,mm$)

TS EAMCET 2015 All TS EAMCET

A
$12.6$
B
$1.27$
C
$2.532$
D
$25.3$

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Wave Optics

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Single Slit Diffraction of Light

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