u (d) 2c v u
9. A glass slab of thickness 8 cm contains the same number of waves as 10 cm of water, when both are traversed by the same monochromatic light. If the
chapterwise McQs for practice
Useful for All National and State Level Medical/Engg. Entrance Exams
refractive index of water is 4/3, the refractive index of glass is
(a) 5/4 (b) 3/2 (c) 5/3 (d) 16/15 10. Assuming the diameter of the eye pupil to be 2.0 mm.
The smallest angular separation at which two point objects can be distinctly seen when viewed in light of wavelength 6000 Å is
(a) 3.66 × 10–4 rad (b) 2.56 × 10–4 rad (c) 1.09 × 10–4 rad (d) 4.51 × 10–3 rad
11. Two coherent light sources A and B are at a distance 3l from each other (l = wavelength). The minimum distance from A on the x-axis at which the interference is constructive is
(a) 3l (b) 4l (c) 5l (d) 8.75l 12. In Fraunhoffer diffraction at a single slit of width
d with incident light of wavelength 5500 Å, the first minimum is observed at angle of 30°. The first secondary maximum is observed at angle q, equal to
(a) sin−
1 1
2 (b) sin−
1 1
4 (c) sin−
1 3
4 (d) sin−
1 3
2 13. Two ideal slits S1 and S2 are at
a distance d apart and illuminated by light of wavelength passing through an ideal source. Slit S placed on the line through S2 as shown in figure. The distance between the planes of slits
and the source slit is D. A screen is held at a distance D from the plane of the slits. The minimum value of d for which there is darkness at O is
(a) 3
lD (b) lD (c) 3lD (d) lD2 2 14. In a double slit experiment, instead of taking slits of
equal widths, one slit is made twice as wide as the
other. Then in the interference pattern
(a) the intensities of both the maxima and the minima increase
(b) the intensity of maxima increases and the minima has zero intensity
(c) the intensity of maxima decreases and that of minima increases
(d) the intensity of maxima decreases and the minima has zero intensity.
15. A beam of unpolarized light of intensity I0 is passed first through a tourmaline crystal A and then through another tourmaline crystal B and oriented so that its principal plane is parallel to that of A. The intensity of final emergent light is I. The value of I is (a) I0/2 (b) I0/4
(c) I0/8 (d) none of these Dual nature of matter anD raDiation 16. A parallel beam of light is incident normally on
a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is (a) 3.2 × 10–8 N (b) 3.2 × 10–7 N
(c) 5.12 × 10–7 N (d) 5.12 × 10–8 N
17. A photon of energy E ejects a photoelectrons from a metal surface whose work function is W0. If this electron enters into a uniform magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r, is given by
(a) 2m W E0 eB
( − ) (b) 2e E W0 mB
( − )
(c) 2m E W0 eB
( − )
(d) 2mW0 eB
18. What wavelength is corresponding to a beam of electrons whose kinetic energy is 100 eV?
(h = 6.6 × 10–34 J sec; 1 eV = 1.6 × 10–19 J;
me = 9.1 × 10–31 kg)
(a) 1.2 Å (b) 2.4 Å (c) 3.6 Å (d) 4.8 Å 19. Find the maximum kinetic energy of photoelectrons
emitted from the surface of lithium of work function 2.39 eV, when exposed with light wave of electric field given by E = E0 sin [1.57 × 107 m–1(ct – x)]
[Given h = 6.63 × 10–34 J s]
(a) 3.11 eV (b) 2.39 eV (c) 1.56 eV (d) 0.71 eV
20. One milli watt of light of wavelength 4560 Å is incident on a cesium surface of work function 1.9 eV.
Given that quantum efficiency of photoelectric emission is 0.5%, Planck’s constant, h = 6.62 × 10–34 Js, velocity of light c = 3 × 108 m s–1, the photoelectric current liberated is
(a) 1.836 × 10–6 A (b) 1.836 × 10–7 A (c) 1.836 × 10–5 A (d) 1.836 × 10–4 A
21. Given that a photon of light of wavelength 10000 Å has an energy equal to 1.243 eV. When light of wavelength 5000 Å and intensity I0 falls on a photoelectric cell, the surface current is 0.40 × 10–6 A and the stopping potential is 1.36 V, then the work function is
(a) 0.43 eV (b) 0.55 eV (c) 1.12 eV (d) 1.53 eV
22. A cathode ray tube contains a pair of parallel metal plates 1.0 cm apart and 3.0 cm long. A narrow horizontal beam of electrons with a velocity of 3 × 107 m s–1 is passed down the tube mid way between the two plates. When the potential difference of 550 V is maintained across the plates, it is found that the electron beam is so deflected that it just strikes the end of one of the plates. Then the specific charge of an electron in C kg–1 is (a) 1.8 × 109 (b) 1.8 × 1011 (c) 3.6 × 1012 (d) 3.6 × 1014
23. Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 nm is incident upon it, a potential of 7.7 volt is required to stop the photoelectrons from reaching the collector plate.
How much potential will be required to stop the photoelectrons when light of wavelength 200 nm is incident upon silver?
(Given hc = 12375 eV Å) (a) 0.5 V (b) 1.5 V (c) 2.35 V (d) 3.85 V
24. Figure represents a graph of kinetic energy of most energetic photoelectrons Kmax (in eV) and frequency u for a metal used as cathode in photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is
(a) 1 × 1014 Hz (b) 1.3 × 1014 Hz (c) 2.3 × 1014 Hz (d) 2.7 × 1014 Hz
25. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of light of wavelength 200 nm. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A × 10Z (where 1 < A < 10).
The value of Z is
(a) 8 (b) 10 (c) 7 (d) 2
26. The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is
(a) four times the initial energy (b) equal to the initial energy (c) twice the initial energy (d) thrice the initial energy.
27. We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm [1 picometre (pm) = 10–12 m], this means that one must be able to resolve a width of say 10 pm. If an electron microscope is used, the minimum electron energy required is about (a) 1.5 keV (b) 15 keV
(c) 150 keV (d) 1.5 MeV
28. In a photoemissive cell, with exciting wavelength l, the fastest electron has speed v. If the exciting wavelength is changed to 3
4l , the speed of the fastest emitted electron will be
(a) less than v(4/3)1/2 (b) v(4/3)1/2
(c) v(3/4)1/2 (d) greater than v(4/3)1/2 29. A source S1 is producing 1015 photons per second of
wavelength 5000 Å. Another source S2 is producing 1.02 × 1015 photons per second of wavelength 5100 Å.
The, (power of S2)/(power of S1) is equal to
(a) 1.00 (b) 1.02 (c) 1.04 (d) 0.98
30. Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions fp = 2.0 eV, fq = 2.5 eV and fr = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is
(a) (b) bright fringe from l2
\ n1l1 = n2l2 first prism, then intensity of polarised light from first prism,
I1 1I0
=2
Using law of Malus,
intensity of polarised light transmitted from second prism is
For missing wavelengths (2 1)
2 2 falling on first plate. Then intensity of polarised light coming from first plate is
I1 = I0/2
When wavelength is changed to 5500 Å, the new fringe width becomes,
12.50 – 0.25 = 12.25 mm
\ New position of 10th order maxima would be y = 12.25 + 10 b′ = 12.25 + 10 × 0.23 = 14.55 mm 7. (a) : Doppler’s shift, d v
c r
l= × = l ω lc
=
r
T c
2p l
= × × ×
× × × × ×
× 7 10 2 22 25 24 60 60 7
6000 3 10
8
8 Å = 0.04 Å
8. (b) : In this situation, we can assume as if both the source and observer are moving towards each other with a speed v
\ As the no. of waves contained in both the medium are same (Given)
′ = +
− = − +
u ( )u ( )(− )u
( )
c v c v
c v c v c v 2
= −
+ − =
− <<
(c v ) ( )
c v cv
c
c cv v c
2 2
2 2
2
2 2 2
2 2
u u
′ = −
u cu
c v2
9. (c) : Let the wavelength of monochromatic light in glass be lg cm, and in water be lw cm.
\ Number of waves in 8 cm of glass = 8 lg, and number of waves in 10 cm of water = 10 lw As the number of waves contained in both the medium are same (Given)
\ 8 10 10
8 5 l l 4
l
g w l
w
= or g = =
Now mg
g
c
=v and mw
w
c
=v
\ m m
ul ul
g w
w g
w g
v
=v = = 5
4 mg =5mw= × =
4 5 4
4 3
5 3
10. (a) : Here, l = 6000 Å = 6 × 10–7 m, D = 2.0 mm = 2.0 × 10–3 m
The limit of resolution of the eye, dq Dl
=1 22. = × ×
× −− = × −
1 22 6 10
2 0 10 3 7 3 66 10 4 .
. . rad.
11. (b) : Let interference be constructive at any point P on X-axis; AP = x
BP= AB2+AP2 = ( )3l2+x2
Path difference =BP AP− = ( )3l2+x2− =x l
For first secondary maximum dsinq=(2 1n+ )l= l= d
Path difference between the waves reaching O from S Dx = Dx1 + Dx2 For darkness at O,
Dx=(2 1n− ) = × −( )
14. (a) : In interference, it is known that
Imax=( I1+ I2) ,2 and Imin=( I1− I2)2 When widths of both the slits are equal, I1 = I2 = I \ Imax = 4I and Imin = 0
When width of one of the slits is increased, intensity due to that slit would increase, while that due to the other will remain the same.
Let I1 = I and I2 = nI where n > 1
\ Imax=I(1+ n)2>4I Imin=I n( −1)2>0
\ Intensities of both, maxima and minima increase.
15. (a) : Intensity of polarised light from tourmaline crystal A I= 0
As principal plane of crystal B is parallel to that of A, 2 therefore, intensity of final emergent light is I = I0/2 cos2 0° = I0/2
16. (b) : Momentum of incident light per second
p E
1 c 608 7
3 10 2 10
= = × = × −
Momentum of reflected light per second
p E
\ Force on the surface = Change in momentum per second
= p2 – (–p1) = p2 + p1
= (1.2 + 2) × 10–7 = 3.2 × 10–7 N 17. (c) : We know that Einstein equation,
E W= 0+1mv2
2 or v E W
= 2( m− 0)
and a charged particle placed in uniform magnetic field experience a force,
F mv= r2 ⇒ evB mv
Comparing it with the equation of harmonic wave in electric field;