Resolving power of a compound microscope:

1.  Depends on the wavelength of light as \(\propto\) $\lambda$

2.  Depends on the wavelength of light as \(\propto\) $\lambda$²

3.  Depends on the wavelength of light as \(\propto\) $\frac{1}{\lambda }$

4.  Depends on the wavelength of light as \(\propto\) $\frac{1}{{\lambda }^2}$

The plane wavefront is incident on a spherical mirror as shown. The reflected wavefront will be:

1.		2.
3.		4.

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 center of the screen is:
1. \(0.10~\text{cm}\)
2. \(0.25~\text{cm}\)
3. \(0.20~\text{cm}\)
4. \(0.15~\text{cm}\)

In Young's double-slit experiment, the separation \(d\) between the slits is \(2\) mm, the wavelength \(\lambda\) of the light used is \(5896~\mathring{A}\) and distance \(D\) between the screen and slits is \(100\) 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. \(1.8\) mm
2. \(1.9\) mm
3. \(2.1\) mm
4. \(1.7\) mm

The intensity at the maximum in a Young's double-slit experiment is \(I_0\). Distance between 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 \(D = 10d\)?
1. \(\frac{I_0}{4}\)
2. \(\frac{3I_0}{4}\)
3. \(\frac{I_0}{2}\)
4. \(I_0\)

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 \(\frac{\lambda}{4}\) will be:
1. \(K\)
2. \(\frac{K}{4}\)
3. \(\frac{K}{2}\)
4. zero