Question 350: Waves Optics

Question

In a diffraction pattern due to single slit of width '$$a$$', the first minimum is observed at an angle of $$30^{\circ}$$ when the light of wavelength $$5400 \mathop A\limits^o$$ is incident on the slit. The first secondary maximum is observed at an angle of $$\left(\sin 30^{\circ}=\frac{1}{2}\right)$$

Accepted Answer

For $$\mathrm{n}^{	ext {th }}$$ secondary minimum,
path difference $$=a \sin 	heta_n=n \lambda$$
For $$\mathrm{n}^{	ext {th }}$$ secondary maximum,
path difference $$=a \sin 	heta_n=(2 n+1) \frac{\lambda}{2}$$
$$	herefore \quad$$ For $$1^{	ext {st }}$$ minimum, a $$\sin 30^{\circ}=\lambda$$ ..... (i)
For $$2^{	ext {nd }}$$ maximum, a $$\sin 	heta_{\mathrm{n}}=(2+1) \frac{\lambda}{2}=\frac{3 \lambda}{2}$$ ..... (ii)
$$	herefore \quad$$ Dividing equation (i) by equation (ii),
$$\frac{\left(\frac{1}{2}ight)}{\sin 	heta_n}=\frac{2}{3} \Rightarrow 	heta_n=\sin ^{-1}\left(\frac{3}{4}ight)
$$

Answer

$$\sin ^{-1}\left(\frac{3}{4}\right)$$

Answer

$$\sin ^{-1}\left(\frac{2}{3}\right)$$

Answer

$$\sin ^{-1}\left(\frac{1}{2}\right)$$

Answer

$$\sin ^{-1}\left(\frac{1}{4}\right)$$

Waves Optics

In a diffraction pattern due to single slit of width '$a', the first minimum is observed at an angle of 30^{\circ} when the light of wavelength 5400 \mathop A\limits^o is incident on the slit. The first secondary maximum is observed at an angle of \left(\sin 30^{\circ}=\frac{1}{2}\right)

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