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. ${\mathrm{I}}_0$

In a diffraction pattern due to a single slit of width a,the first minimum is observed at an angle ${30}^{\circ }$ when light of wavelength 5000 $\dot{A}$ is incident on the slit. The first secondary maximum is observed at an angle of 

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

For a parallel beam of monochromatic light of wavelength 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)2Dλ/a

(2)Dλ/a

(3)Da/λ

(4)2Da/λ

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

1. 0.2 mm

2. 0.1 mm

3. 0.5 mm

4. 0.02 mm