(4)

Here, $a=0.02$ $\text{cm}=2\times {10}^{-4}$ $\text{m}$

$\lambda =5\times {10}^{-5}\text{cm}=5\times {10}^{-7}\text{m} \Rightarrow D=60\text{cm}=0.6\text{m}$

Position of first minima on the diffraction pattern,

$y=\frac{D\lambda }{a}=\frac{0.6\times 5\times {10}^{-7}}{2\times {10}^{-4}}=15\times {10}^{-4}$ $\text{m}=0.15$ $\text{cm}$

(2)

For first minimum, the path difference between extreme waves, $a\sin \theta =\lambda$

Here $\theta =30°\Rightarrow \sin \theta =\frac{1}{2} \therefore a=2\lambda$                    ...(i)

For first secondary maximum, the path difference between extreme waves

$a\sin \theta \text{'}=\frac{3}{2}\lambda \text{or} \left(2\lambda \right)\sin \theta \text{'}=\frac{3}{2}\lambda \text{[Using eqn (i)]}$

or  $\sin \theta \text{'}=\frac{3}{4} \therefore$ $\theta \text{'}={\sin }^{-1}\left(\frac{3}{4}\right)$

(3)

For double slit experiment,

$d=1$ $\text{mm}=1\times {10}^{-3}$ $\text{m}, D=1$ $\text{m}, \lambda =500\times {10}^{-9}$ $\text{m}$

$\text{Fringe width }\beta =\frac{D\lambda }{d}$

$\text{Width of central maxima in a single slit }=\frac{2\lambda D}{a}$

As per question, width of central maxima of single slit pattern = width of 10 maxima of double slit pattern

$\frac{2\lambda D}{a}=10\left(\frac{{\displaystyle \lambda D}}{{\displaystyle d}}\right) \text{or} a=\frac{2d}{10}=\frac{2\times {10}^{-3}}{10}=0.2\times {10}^{-3}\text{m}=0.2$ $\text{mm}$

(1)

The situation is shown in the figure.

[IMAGE 288] ----------------------------------------------------------------

In figure A and B represent the edges of the slit AB of width $a$ and C represents the midpoint of the slit.

For the first minimum at P,

       $a\sin \theta =\lambda$                                                                    ...(i)

where $\lambda$ is the wavelength of light.

The path difference between the wavelets from A to C is

        $\Delta x=\frac{a}{2}\sin \theta =\frac{1}{2}\left(a\sin \theta \right)=\frac{\lambda }{2} \text{(using (i))}$

The corresponding phase difference $\Delta \varphi$ is

        $\Delta \varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }\times \frac{\lambda }{2}=\pi$

(4)

Here, $\lambda =600$ $\text{nm}=600\times {10}^{-9}$ $\text{m}$

$a=1$ $\text{mm}={10}^{-3}$ $\text{m}, D=2$ $\text{m}$

Distance between the first dark fringes on either side of the central bright fringe is also the width of central maximum.

$\text{Width of central maximum}=\frac{2\lambda D}{a}$

$=\frac{2\times 600\times {10}^{-9}\text{m}\times 2\text{m}}{{10}^{-3}\text{m}}=24\times {10}^{-4}\text{m}=2.4\times {10}^{-3}\text{m}=2.4$ $\text{mm}$