The angular resolution of a 10 cm diameter telescope for a wavelength of 5000 Å is of the order of:

1. 10^–4 rad

2. 10^–6 rad

3. 10⁶ rad

4. 10^–2 rad

A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of:

1. 5 m

2. 5 mm

3. 5 cm

4. 0.5 m

The diameter of the human eye lens is 2 mm. What will be the minimum distance between two points to resolve them, which are situated at a distance of 50 meters from the eye? The wavelength of light is 5000 Å:

1. 2.32 m

2. 4.28 mm

3. 1.52 cm

4. 12.48 cm

Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inches?

$$\begin{gathered} 1.<mspace\; width="1em"></mspace>1.2\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{radians} \\ 2.<mspace\; width="1em"></mspace>3.3\times {10}^{-7}<mspace\; width="1em"></mspace>\mathrm{radians} \\ 3.<mspace\; width="1em"></mspace>1.7\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{radians} \\ 4.<mspace\; width="1em"></mspace>2.9\times {10}^{-7}<mspace\; width="1em"></mspace>\mathrm{radians} \end{gathered}$$

For what distance is ray optics a good approximation when the aperture is \(3\) mm wide and the wavelength is \(500\) nm?
1. \(32\) m
2. \(42\) m
3. \(18\) m
4. \(20\) m

Given below are two statements: 

The resolving power of a compound microscope will be maximum when:

Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is:

1. $1.83\times {10}^{-7}$ $rad$

2. $7.32\times {10}^{-7}$ $rad$

3. $6.00\times {10}^{-7}$ $rad$

4. $3.66\times {10}^{-7}$ $rad$

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

1.		2.
3.		4.