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Physics NEET MCQ
Quiz 9
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Q.1
In Young's double slit experiment, if L is the distance between the slits and the screen upto which interference pattern is observed, x is the average distance between the adjacent fringes and d being the slit separation. The wave length of light is given by [ MPPMT 1993]
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a)xd/L
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b) xL/d
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c)Ld/x
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d)1/Ldx
Explanation
USe formula for fringe widthAnswer: (a)
Q.2
Two phases related and monochromatic beams of light have intensities I and 4I. Possible maximum and minimum intensities in the resultant beam obtained due to superposition are [ MPPMT 1999]
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a) 5I and I
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b) 5I and 3I
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c)9I and 3I
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d)9I and I
Explanation
If E is amplitude then I ∝ E2 or √I ∝ E and 2√I ∝ E' For maxima Emax=E +E'Maximum intensity ∝ (E+E')2 Maximum intensity ∝ (√I +2√I )2=9I For minimum Emin=E'-E minimum intensity ∝ (E'- E)2 Maximum intensity ∝ (2√I -√I )2=I Answer: (d)
Q.3
In Young's double slit experiment, the separation between the slits is halved and the distance between the slit and screen is doubled. The fringe-width will [ CPMT 1998]
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a) remain unchanged
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b) be halved
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c)be doubled
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d)be four times
Explanation
USe formula for fringe width w - λD/ d d=distance between the slits and D=distance between the slit and screen Answer:(d)
Q.4
A ray of light is incident on the surface of a glass plate at an angle φ. If µ represents the refractive index of glass with respect to air, then the angle between the reflected and refracted ray is [ CPMT 1989]
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a) 90° + φ
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b) sin⁻¹(µcosφ)
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c) 90°
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d) 90° - sin⁻¹ ( sinφ/µ)
Explanation
Answer: (c)
Q.5
In a certain double slit experimental arrangement. Interference fringes of width 1.0 mm each are observed when light of wavelength 5000Å is used. Keeping the set-up unaltered, if the source is replaced by another of wavelength 6000Å, the fringe width will be [ CPMT 1988]
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a)0.5 mm
0%
b) 1.0 mm
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c)1.2 mm
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d)1.5 mm
Explanation
We know that fringe width w ∝ λ take ratio form two wave lengths Answer: (c)
Q.6
When light wave suffers reflection at the interface from air to glass, the change in phase of the reflected wave equal to [ CPMT 1991]
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a) 0
0%
b) π/2
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c)π
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d)2π
Explanation
Answer: (c)
Q.7
A Young's double slit set-up for interference is shifted from air to within water, then the fringe width [ RajPMT 1997]
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a) Becomes infinite
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b) Decreases
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c)Increases
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d)Remains unchanged
Explanation
we know that refractive index µ=λair /λmedium Thus wavelength decreases in transparent mediumAlso fringe width w ∝ wavelength Answer:(b)
Q.8
Plane polarized light is incident on an analyser. The intensity then becomes three fourth. The angle of the axis of the analyzer with the beam is
0%
a)30°
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b) 45°
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c)60°
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d)zero
Explanation
I=Iocos2θ (3/4)Io=I0cos2θcosθ=√3 /2 θ=60° This is the angle, which the axis of analyzer makes with vertical. Therefore the angle which the axis of analyzer makes with light ray=90 - 60=30°Answer: (c)
Q.9
In Young's double slit experiment when wavelength used is 6000Å and the screen is 40cm from the slits, the fringes are 0.012cm apart. What is the distance between the slits? [ PMT 1995]
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a) 0.024cm
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b) 2.4cm
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c)0.24cm
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d)0.2cm
Explanation
use formula for fringe widthAnswer: (d)
Q.10
In Young's double slit experiment, the slits are 0.5 mm apart and interference pattern is observed on a screen placed at a distance of 1.0 m from the plane containing the slits. If wavelength of the incident light is 6000Å then the separation between the third bright fringe and the central maxima is [ AMU 1995]
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a) 4.0 mm
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b) 3.6 mm
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c)3.0 mm
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d)2.5 mm
Explanation
USe formula for position of nth bright fringe=nλD /d Answer:(b)
Q.11
A slit of width a is illuminated by white light. The first minimum for red light ( λ=6500Å) will fall at θ=30° when a will be [ MPPMT 1987]
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a) 3250Å
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c) 1.3 micron
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b) 6.5×10⁻⁴ mm
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d) 2.6×10⁻⁴ cm
Explanation
from the formula for minimum of diffraction a sinθ=nλ Answer: (c)
Q.12
A diffraction pattern is obtained using a beam of red light. What happens if red light is replaced by the blue light [ CPMT 1998]
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a)no change
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b) diffraction bands become narrow and crowded together
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c)diffraction bands become broader and farther apart bands disappear
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d)bands disappear
Explanation
width ∝ wavelengthAnswer: (b)
Q.13
Ray optics is valid, when characteristic dimensions are [ CBSE 1994]
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a) of much order as the wavelength of light
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b) much smaller than the wavelength of light
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c)of the order of one millimeter
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d)much larger than the wavelength of light
Explanation
Answer: (d)
Q.14
Light travels in straight line because [ Raj.PMT 1997]
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a) It is not absorbed by atmosphere
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b) Its velocity is very high
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c)Diffraction effect is negligible
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d)None of these
Explanation
Answer:(c)
Q.15
A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in figure. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. the rays AB and A'B' undergo interference. The ratio Imax / I min is : [ IIT 1990]
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a) 4 : 1
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b) 8 : 1
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c) 7 : 1
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d) 49 : 1
Explanation
Intensity of reflected ray AB=0.25I=I/4 Intensity of refracted ray AC=0.75I=3I/4Intensity of reflected ray CA'=(0.25)(3I/4)=3I/16 Intensity of refracted ray A'B'=0.75(3I/16)=9I/64 Answer: (d)
Q.16
Young's experiment establish the fact that [ MPPET 1994]
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a)Light consists of particles
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b) Light consists of waves
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c)Light consists of neither particles nor waves
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d)fringe width do not depend on the separation of slits
Explanation
Answer: (b)
Q.17
Light wave travel in vacuum along the x-axis which of the following represent the wavefronts?
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a) x=a
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b) y=a
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c) z=a
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d) x+y+z=a
Explanation
Wave front is in Y-Z plane thus values of x must be constant Answer: (a)
Q.18
A ray falls on a prism ABC ( AB=BC) and travels as shown in figure. The minimum refractive index of the prism material should be
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a) 4/3
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b) √2
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c) 1.5
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d) √3
Explanation
Answer: (b)
Q.19
White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and the screen is at distance d( d > b) from the slits. At a point on the screen directly in front of one of the slits certain wavelengths are missing. One of the missing wave length is [ IIT 1984]
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b) λ=2b2 / d
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c)λ=3b2 /d
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d)λ=2b2 /3d
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a) λ=b² / d
Explanation
For certain wavelength missing, interference should be destructive for that wavelengthExpanding binomially and neglecting higher powers Path difference=d+ (b2 / 2d) - d=(b2 / 2d) For distractive interference Answer:(a)
Q.20
Casting of geometrical shadows is due to the phenomenon of [ CPMT 1999]
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a) Diffraction
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b) Polarization
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c) Interference
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d) Refraction
Explanation
Answer: (a)
Q.21
In an interference pattern produced by two identical slits, the intensity at the site of the central maximum is I. The intensity at same spot when either of two slits is closed to IWe must have [ BHU 1998]
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a) I=Io
0%
b) I=2Io
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c)I=4Io
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d)I and Ioare not related
Explanation
Intensity due to one slit=Io∴ Due to two slits it will be=(√Io + √Io) 2 Or I=4IoAnswer: (c)
Q.22
Two points separated by a distance of 0.1 mm can just be inspected in a microscope when a wavelength 6000Å is used. If the light of wavelength 4800Å is used this limit of resolution will become [ CPMT 1988]
0%
a) 0.8 mm
0%
b) 0.12 mm
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c)0.10 mm
0%
d)0.08 mm
Explanation
limit of resolution ∝ λ Answer:(d)
Q.23
The resolution limit of eye is 1'. At a distance of X km from eye, two persons standing with lateral separation of 3 m. For the two persons to be just resolved by the naked eye, X should be [ CPMT 1989]
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a) 10 km
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b) 15 km
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c) 20 km
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d) 30 km
Explanation
1'=π / 10800 resolution=separation / distance Answer: (a)
Q.24
The phenomenon of interference is shown by [ PMPET 1997]
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a)Longitudinal mechanical wave only
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b) Transverse mechanical wave only
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c)Non mechanical transverse wave only
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d)All the above types of waves
Explanation
Answer: (d)
Q.25
The intensity ratio I1 / I2 of the two interfering sources in Young's experiment isThe ratio Imax / Imin is [ BHU 1995]
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a) 4:1
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b) 2:1
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c)3:1
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d)9:1
Explanation
Answer: (d)
Q.26
In Young's two slit interference experiment the distance between the slits is made 3 fold the fringewidth becomes [ CPMT 1989]
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a) 1/3 fold
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b) 3 fold
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c)1/9 fold
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d)9 fold
Explanation
Answer:(a)
Q.27
In Young's experiment, the sodium lamp of wavelength λ=5898Å produces 92 fringes in visible region, if the source of light is changed by green light of wavelength 5461Å the number of fringes obtained in the visible region will be [ Rj.PET 1996]
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a) 62
0%
b) 67
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c) 85
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d) 99
Explanation
Field view is same so 92×5898=n×5461 n=99 Answer: (d)
Q.28
Wave front means [ Raj.PMT 1997]
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a)All particles in it have same phase
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b) All particles have opposite phase of vibration
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c)Few particles are in same phase, rest are in opposite phase
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d)None of these
Explanation
Answer: (a)
Q.29
In Young's double slit experiment angular width of fringes is 0.20° for sodium light of wave length 5890Å. If complete system is dipped in water then angular width of fringes becomes [ Raj.PET 1997]
0%
a) 0.11°
0%
b) 0.15°
0%
c)0.22°
0%
d)0.30°
Explanation
θ=λ/d θ1 / θ2=λ1 / λ2 λ1 / λ2=µ ∴ θ1 / θ2=µ θ2=θ1 / µθ2=0.2/×3 / 4=0.15°Answer: (b)
Q.30
Plane polarized light is passed through a polaroid. On viewing through the polaroid we find that when the polaroid is given one complete rotation about the direction of the light, one of the following is observed [ MNR 1993]
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a) The intensity of light gradually decreases to zero and remains at zero
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b) The intensity of light gradually increases to maximum and remains at maximum
0%
c)There is no change in intensity
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d)The intensity of light is twice maximum and twice zero
Explanation
Answer:(d)
Q.31
In Young's experiment the intensities of bright and dark fringes are 4I and I respectively, the ratio of the amplitudes of two waves will be [ Raj.PMT 1996]
0%
a) 1:1
0%
b) 1:4
0%
c) 3:1
0%
d) 1:2
Explanation
Let E1 and E2 be the amplitude of the two waves Answer: (c)
Q.32
Light appears to travel in straight lines since [ CPMT 1990]
0%
a)It is not absorbed by the atmosphere
0%
b) It is reflected by the atmosphere
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c)Its wave length is very small
0%
d)Its velocity is very large
Explanation
Answer: (c)
Q.33
Consider Fraunhofer diffraction pattern obtained with a single slit, illuminated at normal incidence. At the angular position of the first diffraction minimum, the phase difference ( in radians ) between the wavelets from the opposite edges of the slit is [ IIT 1995]
0%
a) π
0%
b) 2π
0%
c)π/4
0%
d)π/2
Explanation
First minima in single slit experiment lies between +π and -π Therefore the required answer is=π - ( -π)=2πAnswer: (b)
Q.34
In double slit experiment, the angular width of the fringes is 0.2° for sodium light ( λ=5890Å ) . In order to increase the fringe width by 10% the necessary change in the wavelength is [ MPPMT 1997]
0%
a) increase of 589Å
0%
b) decrease of 589Å
0%
c)increase of 6479Å
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d)zero
Explanation
We know that θ=λ/d θ2=0.22° λ2 - λ1=589Å Answer:(a)
Q.35
A beam of light AO is incident on a glass slab ( µ=1.54) in a direction as shown in the adjoining diagram. The reflected ray OB is passed through a Nicol prism. On viewing through the Nicol prism, we find on rotating the prism that ( given tan57°=1.54) [ CPMT 1986]
0%
a) The intensity is reduced down ro zero and remains zero
0%
b) The intensity reduces down some what and rises again
0%
c) there is no change in intensity
0%
d) The intensity gradually reduces to zero and then again increases
Explanation
Angle of incidence is 57° According to Brewester law if tani=µ then reflected ray's are plane polarized. In given problem condition is satisfied thus reflected ray's are plane polarized Answer: (d)
Q.36
In double slit experiment, for light of which colour the fringe width will be minimum [ MPPMT 1994]
0%
a)Violet
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b) Red
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c)Green
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d)Yellow
Explanation
Answer: (a)
Q.37
While both light and sound shows wave character, diffraction is much higher to observe in light. This is because [ CET 1994]
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a) light does not require a medium
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b) wavelength of light is far smaller
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c)waves of light are transverse
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d)speed of light is far greater
Explanation
Answer: (b)
Q.38
Find the thickness of plate which will produce a change in optical path equal to half the wavelength Λ of light passing through it, normally. The refractive index of the plate is µ [ CPMT 1998]
0%
a)λ / [4(µ - 1)]
0%
b) 2λ / [4(µ - 1)]
0%
c)λ / (µ - 1)
0%
d)λ / [2(µ - 1)]
Explanation
path difference=(µ - 1) t λ/2=(µ - 1) t t=λ/[2(µ - 1)] Answer:(d)
Q.39
In the adjoining diagram, wavefront AB, moving in air is incident on a plane glass surface XY. Its position CD after refraction through a glass slab is shown also along with the normal's drawn at A and D. The refractive index of glass with respect to air (µ ) will be equal to{( CPMT 1988]
0%
a) sinθ/sinθ'
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b) sinθ/sinφ'
0%
c) sinφ'/sinθ
0%
d) AB /CD
Explanation
From geometry of figure and Snell's law Answer: (b)
Q.40
A thin oil layer floats on water. A ray of light making an angle of incidence of 40° shines on oil layer. The angle of refraction of ray with water surface is ( µoil=1.45, µwater=1.33) [MPPMT 1993]
0%
a)36.1°
0%
b) 44.5°
0%
c)26.8 °
0%
d)28.9°
Explanation
from figure 1.47=sin40 / sinr1 sinr1=0642 / 1.45 r1=0.642 / 1.45 r1=26.31° 1.33/1.45=sinr1 / sin r2 r2=29°Answer: (d)
Q.41
Four different independent waves are represented by i) y1=a1sinωtii) y2=a2sin2ωtiii) y3=a3cosωtiv) y4=a4sin(ωt+π/3) with which of two waves, interference is possible
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a) in (i) and (ii)
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b) in (i) and (iv)
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c)in (iii) and (iv)
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d)not possible with any combination
Explanation
Answer: (d)
Q.42
The refracting angle of prism is A and refractive index of material of prism is cot(A/2). The angle of minimum deviation is
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a) (180° - 3A)
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b) (180° + 2A)
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c)( 90° - A)
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d)(180° - 2A)
Explanation
Answer:(d)
Q.43
A disc is placed on a surface of pond which has refractive index 5/A source of light is placed 4m below the surface of the liquid. The minimum radius of disc needed so that light is not coming out is [ CBSE 2001]
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a)∞
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b) 3m
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c)6 m
0%
d)4 m
Explanation
For total internal reflection θ must be equal to critical angle µ=1 / sinC From figure Answer: (b)
Q.44
A camera objective has an aperture diameter d. If the aperture is reduced to diameter d/2, the exposure time under identical condition of light should be made [ KErala, PMT 2004]
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a) √2 fold
0%
b) 2 fold
0%
c)2√2 fold
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d)4 fold
Explanation
When aperture is reduced to half area becomes 1/4th. Therefore, exposure time should be made 4 foldAnswer: (d)
Q.45
In a compound microscope, the objective lens of fo and eye piece of fe are placed at distance L such that L equals [ kerala P.M.T 2004]
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a)fo + fe
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b) fo - fe
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c)much greater than fo or fe
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d)much less than fo or fe
Explanation
Answer:(a)
Q.46
The dispersive power of the material of lens of focal length 20cm is 0.The longitudinal chromatic aberration of the lens is [ kerala PMT 2004]
0%
a)0.08 cm
0%
b) 1.6 cm
0%
c)0.08/20 cm
0%
d)0.16 cm
Explanation
Longitudinal chromatic aberration=ωf=0.08×20=1.6 cmAnswer: (b)
Q.47
A ray of light passes through an equilateral prism such that angle of incidence is equal to the angle of emergence and the latter is equal to 3/4th angle of prism. The angle of deviation is [ Kerala PET 2004]
0%
a) 45°
0%
b) 39°
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c)20°
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d)30°
Explanation
When i=e, prism is in minimum deviation position i=e=(3/4)A=(3/4)×60°=45° δm=2i - A δm=2×45 - 60=30°Answer: (d)
Q.48
In a compound microscope, the intermediate image is [ kerala PEt 2004]
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a) virtual, erect and magnified
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b) real, erect and magnified
0%
c)real, inverted and magnified
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d)virtual, ere and magnified
Explanation
Answer:(c)
Q.49
A fish at depth of 12cn in water is viewed by an observer on the bank of lake. Through what height is the image of fish raised [µ=4/3) )kera PET 2004]
0%
a) 9 cm
0%
b) 12 cm
0%
c) 3 cm
0%
d) 3.8 cm
Explanation
µ=real depth / apparent depth 4/3 12/ x x=9 cm The height through which image of fish is raised=12 - 9=3 cm Answer: (c)
Q.50
Two plane mirrors are arranged at right angles to each other as shown in figure. A ray of light is incident on the horizontal mirror at an angle Θ. For what value of θ the ray emerges parallel to the incoming ray after reflection from the vertical mirror?
0%
a)75°
0%
b) 45°
0%
c) 30°
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d)all of the above
Explanation
Answer: (d)
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