⇒Wave Optics

Correct Option is (B)

The distance between the central bright fringe and the first dark fringe is given as:

$\begin{array}{rl}{y}_{n_d}& =\frac{n\lambda D}{d}\\ y_{1_d}& =\frac{1\times 600\times {10}^{-9}\times 2}{{10}^{-3}}\\ & =1.2\times {10}^{-3}=1.2\mathrm{mm}\end{array}$

$\therefore$ The distance between the first dark fringe on either side of the central bright fringe is $2{\mathrm{y}}_{\mathrm{n}}=2\times 1.2=2.4\mathrm{mm}$

Correct Option is (C)

Width of the central maxima $=W=\frac{\lambda \theta }{d}$

Thus, the slit width is inversely proportional to the width of the central maximum.

Hence, when the slit width is decreased, the width of the central maxima increases.

Correct Option is (C)

Width of central maximum ${\mathrm{W}}_{\mathrm{c}}=2\left[\frac{\lambda \mathrm{D}}{\mathrm{a}}\right]$

$\Rightarrow {\mathrm{W}}_{\mathrm{c}}\propto \frac{1}{\mathrm{a}}$

$\therefore$ Angular width of principal maximum decreases with increase in width of the slit.

Correct Option is (B)

In diffraction of light by single slit, the width of central maximum is given as

${\mathrm{W}}_{\mathrm{c}}=\frac{2\lambda \mathrm{D}}{\mathrm{d}}$

Given: ${\mathrm{W}}_{\mathrm{c}}=\mathrm{d}$

$\begin{array}{rl}\therefore d& =\frac{2\lambda D}{d}\\ \Rightarrow D& =\frac{d^2}{2\lambda }\end{array}$

Correct Option is (A)

For ${\mathrm{n}}^{\text{th }}$ secondary minimum,

path difference $=a\sin {\theta }_n=n\lambda$

For ${\mathrm{n}}^{\text{th }}$ secondary maximum,

path difference $=a\sin {\theta }_n=(2n+1)\frac{\lambda }{2}$

$\therefore$ For $1^{\text{st }}$ minimum, a $\sin {30}^{\circ }=\lambda$ ..... (i)

For $2^{\text{nd }}$ maximum, a $\sin {\theta }_{\mathrm{n}}=(2+1)\frac{\lambda }{2}=\frac{3\lambda }{2}$ ..... (ii)

$\therefore$ Dividing equation (i) by equation (ii),

$\frac{\left(\frac{1}{2}\right)}{\sin {\theta }_n}=\frac{2}{3}\Rightarrow {\theta }_n={\sin }^{-1}\left(\frac{3}{4}\right)$