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Physics Question on Wave optics

Light of wavelength 5000 ?5000 \,? is incident normally on a slit of width 2.5×10−4 cm2.5 \times 10^{-4} \, cm. The angular position of second minimum from the central maximum is

A

sin⁡−1(15)\sin^{-1} \left(\frac{1}{5}\right)

B

sin⁡−1(25)\sin^{-1} \left(\frac{2}{5}\right)

C

(π3)\left(\frac{\pi}{3}\right)

D

(π6)\left(\frac{\pi}{6}\right)

Answer

sin⁡−1(25)\sin^{-1} \left(\frac{2}{5}\right)

Explanation

Solution

We know that,
Angular width of central maximum =2λa=\frac{2 \lambda}{a}
where, λ=\lambda= wavelength of light,
a=a= width of single slit,
So, sin⁡θ=2λa\sin \theta=\frac{2 \lambda}{a}
where λ=5000 ?=5000×10−10 m \lambda =5000\, ?=5000 \times 10^{-10}\, m
a=2.5×10−6ma =2.5 \times 10^{-6} m
⇒sin⁡θ=2×5000×10−102.5×10−6\Rightarrow \sin \theta =\frac{2 \times 5000 \times 10^{-10}}{2.5 \times 10^{-6}}
sin⁡ θ=100000×10−1025×10−6\sin \,\theta =\frac{100000 \times 10^{-10}}{25 \times 10^{-6}}
sin⁡ θ=10×1010×10−1025\sin \,\theta =\frac{10 \times 10^{10} \times 10^{-10}}{25}
sin⁡ θ=1025\sin \,\theta =\frac{10}{25}
θ=sin⁡−1(25)\theta =\sin ^{-1}\left(\frac{2}{5}\right)