In Young's double-slit experiment, the separation $d$ between the slits is $2~\text{mm}$, the wavelength $\lambda$ of the light used is $5896~\mathring{A}$ and distance $D$ between the screen and slits is $100~\text{cm}$. It is found that the angular width of the fringes is $0.20^{\circ}$. To increase the fringe angular width to $0.21^{\circ}$ (with same $\lambda$ and $D$) the separation between the slits needs to be changed to:
1. $1.8~\text{mm}$
2. $1.9~\text{mm}$
3. $2.1~\text{mm}$
4. $1.7~\text{mm}$

Two polaroids $P_1$ and $P_2$ are placed with their axis perpendicular to each other. Unpolarised light of intensity $I_0$ is incident on $P_1$. A third polaroid $P_3$ is kept in between $P_1$ and $P_2$ such that its axis makes an angle $45^\circ$ with that of $P_1$. The intensity of transmitted light through $P_2$ is:
1. $\dfrac{I_0}{4}$

2. $\dfrac{I_0}{8}$

3. $\dfrac{I_0}{16}$

4. $\dfrac{I_0}{2}$

A linear aperture whose width is $0.02$ cm is placed immediately in front of a lens of focal length $60$ cm. The aperture is illuminated normally by a parallel beam of wavelength $5\times 10^{-5}$ cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is:

The intensity at the maximum in Young's double-slit experiment is $I_0$. The distance between the two slits is  $d= 5\lambda$,  where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10 d$?
1. $\dfrac{I_0}{4}$

2. $\dfrac{3}{4}I_0$

3. $\dfrac{I_0}{2}$

4. $I_0$