An astronaut is looking down on earth's surface from a space shuttle at an altitude of

$400 km$. Assuming that the astronaut's pupil diameter is 5 mm and the wavelength of

visible light is 500 nm. The astronaut will be able to resolve linear object of the size of

about 

1. 0.5 m                   

2. 5 m

3. 50 m                     

4. 500 m

The distance of the moon from earth is $3.8\times {10}^5 km$. The eye is most sensitive to light of

wavelength 5500 Å. The minimum separation between two points on the moon that can

be resolved by a 500 cm telescope will be 

1. 51 m                         

2. 60 m

3. 70 m                         

4. All the above

A light source is located at \(P_1\) as shown in the figure. All sides of the polygon are equal. The intensity of illumination at \(P_2\) is \(I_0.\)What will be the intensity of illumination at \(P_3?\)

1. \(\dfrac{3}{4} I_{0}\)                                   

2. \(\dfrac{I_{0}}{8}\)

3. \(\dfrac{3}{8} I_{0}\)                                       

4. \(\dfrac{\sqrt{3}}{8} I_{0}\)

We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm, this

means that one must be able to resolve a width of say 10 p.m. If an electron

microscope is used, the minimum electron energy required is about

1. 1.5 KeV                         

2. 15 KeV

3. 150 KeV                         

4. 1.5 KeV

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. 0.5 m                        

2. 5 m

3. 5 mm                       

4. 5 cm

Two point white dots are 1mm apart on a black paper. They are viewed by eye of pupil

diameter 3 mm. Approximately, what is the maximum distance at which dots can be

resolved by the eye ? [Take wavelength of light = 500 nm]

1. 6 m                           

2. 3 m

3. 5 m                           

4. 1 m

The ratio of resolving powers of an optical microscope for two wavelengths ${\lambda }_1=4000 \stackrel{o}{A } and {\lambda }_2=6000 \stackrel{o}{A}$ is 

1. 8:27

2. 9:4

3. 3:2

4. 16:81

Young's double-slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in the air. The refractive index of the medium is nearly 

1. 1.25 

2. 1.59 

3. 1.69

4. 1.78 

Two polaroids \(P_{1}\) and \(P_{2}\) are placed with their axis perpendicular to each other. Unpolarised light \(I_{o}\) 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 \(\left(45\right)^{\circ}\) with that of \(P_{1}\). The intensity of transmitted light through \(P_{2}\)

1. \(\frac{I_{o}}{2}\)

2. \(\frac{I_{o}}{4}\)

3. \(\frac{I_{o}}{8}\)

4. \(\frac{I_{o}}{16}\)

The intensity at the maximum in Young's double-slit experiment is ${\mathrm{I}}_0$ when the 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= 10 d?

1. $\frac{{\mathrm{I}}_0}{4}$               

2. $\frac{3}{4}{\mathrm{I}}_0$

3. $\frac{{\mathrm{I}}_0}{2}$             

4. ${\mathrm{I}}_0$