A screen is placed 100 cm from a single slit which is illuminated with 5000 $\stackrel{\mathrm{o}}{\mathrm{A}}$ light. If the distance between the first and third minima in the diffraction pattern is 5 mm. The width of the slit is

1.  2 mm

2.  0.2 $\mu$m

3.  $2<mspace\; width="1em"></mspace>\times {10}^{-4}<mspace\; width="1em"></mspace>\mathrm{m}$

4.  $4<mspace\; width="1em"></mspace>\times {10}^{-5}<mspace\; width="1em"></mspace>\mathrm{m}$

When the light of a certain wavelength is incident on a plane surface of a material of a glancing angle $30°$, the reflected light is polarised. The angle of refraction is

1.  $60°$

2.  $30°$

3.  $45°$

4.  $40°$

The angular width of the principal maximum in Fraunhofer single slit diffraction is 0.1 radian. The angular width of second-order secondary maxima is

1.  0.05 radian

2.  0.1 radian

3.  0.5 radian

4.  0.25 radian

The resolving power of a microscope can be increased by using:

1.  red light.
2.  blue light.
3.  oil between objective lens and object.
4.  both (2) and (3).

Assume that light of wavelength 7000 $\stackrel{\mathrm{o}}{\mathrm{A}}$ is coming from a star. The limit of resolution (in radian) of a telescope whose objective has a diameter of 244 cm, will be

1.  $2.5<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-7}$

2.  $5.0<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-7}$

3.  $3.5<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-7}$

4.  $4.5<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-7}$

Yellow light is used in a single slit diffraction pattern with a slit width of 0.5 mm. If the yellow light is replaced by X-rays, then the observed pattern will reveal

1.  No diffraction pattern

2.  More narrow central maximum

3.  More number of fringes

4.  Less number of fringes

A diffraction pattern is obtained by using a beam of red light. What will happen, if the red light is replaced by blue light?

1.  Bands will become narrower

2.  Bands become broader

3.  No change will take place

4.  Bands disappear

When a thin transparent plate of thickness t and refractive index $\mu$ is placed in the path of one of the two interfering waves of light, then the extra path difference created between two waves due to the plate is

1.  $\left(\mu <mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>1\right)t$

2.  $\left(\mu <mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>1\right)t$

3.  $\left(\frac{\mu +1}{t}\right)$

4.  $\mu$t

Huygens' wave theory allows us to know the:

The graph between resolving power and accelerating potential V for an electron microscope is (P is resolving power):

1.		2.
3.		4.