In Young's double-slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths ${\lambda }_1$ = 12000 Å and ${\lambda }_2$ = 10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

1.  8 mm

2.  6 mm

3.  4 mm

4.  3 mm

In Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is K, ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda$/4 will be

1.  K

2.  K/4

3.  K/2

4.  zero

For a parallel beam of monochromatic light of wavelength '$\mathrm{\lambda }$', diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the width of the central maxima will be

1.  $\frac{\mathrm{D\lambda }}{\mathrm{a}}$

2.  $\frac{\mathrm{Da}}{\mathrm{\lambda }}$

3.  $\frac{2\mathrm{Da}}{\mathrm{\lambda }}$

4.  $\frac{2\mathrm{D\lambda }}{\mathrm{a}}$

In a double-slit experiment, the two slits are 1 mm apart and the screen is placed I m away A monochromatic light of wave. length 500 nm is used. What will be the width of each slit for obtaining 10 maxima of double-slit within the central maxima of a single-slit pattern?

1.  0.1 mm

2.  0.5 mm

3.  0.02 mm

4.  0.2 mm

Two slits in Young's experiment have a width in the ratio 1:25. The ratio of intensity t the maxima and minima in the interference pattern $\frac{I_{max}}{I_{min}}$ is

1.  $\frac{4}{9}$

2.  $\frac{9}{4}$

3.  $\frac{121}{49}$

4.  $\frac{40}{121}$

In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle $30°$ when the light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of

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

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

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

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

The intensity of the maximum in Young's double-slit experiment is $I_0$. Distance between two slits is d = 5$\mathrm{\lambda }$, where $\mathrm{\lambda }$ 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 = 10d?

1.  $\frac{3}{4}{I}_0$

2.  $\frac{{\mathrm{I}}_0}{2}$

3.  ${\mathrm{I}}_0$

4.  $\frac{{\mathrm{I}}_0}{4}$

The interference pattern is obtained with, two coherent light sources of intensity ratio n. In the interference pattern, the ratio $\frac{I_{max}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>I_{min}}{I_{max}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>I_{min}}$ will be

1.  $\frac{2\sqrt{n}}{n<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>1}$

2.  $\frac{\sqrt{n}}{{\left(n<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>1\right)}^2}$

3.  $\frac{2\sqrt{n}}{{\left(n<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>1\right)}^2}$

4.  $\frac{\sqrt{n}}{n<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>1}$

Two polaroids ${\mathrm{P}}_1$ and ${\mathrm{P}}_2$are placed With their axis perpendicular
to each other. Unpolarised light ${\mathrm{I}}_0$ is incident on ${\mathrm{P}}_1$. A third polaroid ${\mathrm{P}}_3$ is kept in between ${\mathrm{P}}_1$ and ${\mathrm{P}}_2$ such that its axis makes an angle $45°$ with $P_1$ .The intensity of transmitted light through $P_2$ is

1.  $\frac{{\mathrm{I}}_0}{4}$

2.  $\frac{{\mathrm{I}}_0}{8}$

3.  $\frac{{\mathrm{I}}_0}{16}$

4.  $\frac{{\mathrm{I}}_0}{2}$

Unpolarised light is incident from the air on a plane surface of a material of refractive index $\text{'}\mu \text{'}$. At a particular angle of incident 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this statement?

1.  $i = {\sin }^{-1}\left(\frac{1}{\mu }\right)$

2.  The reflected light is polarised with its electric vector perpendicular to the plane of incidence

3.  The reflected light is polarised with its electric vector parallel to the plane of incidence

4.  $i = {\tan }^{-1}\left(\frac{1}{\mu }\right)$