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Ray And Wave Optics Mcq
Quiz 9
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Q.1
The average distance between the earth and moon is 38.6×104km. The minimum separation between the two points on the surface of moon that can be resolved by a telescope whose objective lens has a diameter of 5 metres with λ=6000 Angstrom, is [ MPPMT 1993]
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a)5.65 m
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b) 28.25 m
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c) 11.30 m
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d) 46.51 m
Explanation
Resolving limit of a telescope=λ/d=6×10-7 / 5 --eq(1)Resolving limit=separation / average distance between moon and earth --eq(2) 6×10-7 / 5=x/ 38.6×107 x=38.6×107×1.2×10-7=46.32 mAnswer: (d)
Q.2
An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and eye -piece is 36 cm and the final image is formed at infinity. the focal length fo of the objective and fe of the eye-piece are [ IIT 1989]
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a) fo=45 cm and fe=-9 cm
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b) fo=50 cm and fe=10 cm
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c)fo=7.2 cm and fe=5 cm
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d)fo=30 cm and fe=6 cm
Explanation
Answer: (d)
Q.3
An achromatic combination is to be obtained using a convex and concave lens. The two lenses chosen should have [ MPPMT 1988]
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a) Their powers equal
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b) Their refractive indices equal
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c)Their dispersive power equal
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d)The product of their powers and dispersive powers equal
Explanation
Answer:(d)
Q.4
A source emits light of wavelengths 4700Å, 5400Å and 6500Å. The light passes through red glass before being tested by a spectrometer. Which wavelength is seen in spectrum? [ PMT 1995]
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a) 6500 Å
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b) 6400 Å
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c) 4700 Å
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d) All the three above
Explanation
Only red light will be observed in the spectrum because red glass will transmit only red light Answer: (a)
Q.5
A person can not see the objects clearly placed at distances more than 40 cm. He is advised to use lens of power [ CPMT 1993]
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a)-2.5D
0%
b) +2.5D
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c)-6.25D
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d)+1.5D
Explanation
Answer: (a)
Q.6
Spherical aberration in a thin lens can be reduced by [ IIT 1994]
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a) using a monochromatic light
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b) using a doublet combination
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c)using a circular annular mask over the lens
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d)increasing the size of the lens
Explanation
Answer: (c)
Q.7
The focal length of the objective and eyepiece of a telescope are 100cm and 5cm respectively. Final image is formed at least distance of distinct vision. The magnification of telescope is [ Raj.PET 1997]
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a) 20
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b) 24
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c)30
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d)36
Explanation
Answer:(b)
Q.8
The magnification produced by the objective lens and the eye lens of a compound microscope are 25 and 6 respectively. the magnifying power of this microscope is [ UGET 1995]
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a) 19
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b) 31
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c) 150
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d) √150
Explanation
Answer: (c)
Q.9
A locality is photographed from an aeroplane flying at height of 2000 m with a camera whose focal length is 50cm. The ratio of linear size of the image to the size of the object, is nearly
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a)1 : 10000
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b) 1:40
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c)1:4000
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d)1:1000
Explanation
image will be formed at focal point thus v=50cm u=2000 m m=v/u=50 / 200000=1/4000Answer: (c)
Q.10
In a compound microscope, the focal lengths of objective and eye-lenses are 1.2 cm and 3cm respectively. If the object is put 1.25cm away from the objective lens and the final image is formed at infinity, the magnifying power of the microscope is
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a) 150
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b) 200
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c)250
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d)400
Explanation
Answer: (b)
Q.11
In Young's experiment, monochromatic light is used to illuminate the two slits A and B. Interference fringes are observed on a screen placed in front of the slits. Now if a thin glass glass plate is placed normally in the pth of beam coming from the slit A, then [ MNR 1993]
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a) The fringes will disappear
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b) The fringe width will increase
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c) The fringe width will decrease
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d) There will be no change in fringe width
Explanation
Answer: (d)
Q.12
A ray of light consisting of two wavelengths 4000Å and 5000Å falls from air on a quartz surface. the angle of incidence is 30°, and the indices of refraction for two wave lengths are respectively 1.47 and 1.The angle between the two refracted beams will be [ MPPMT 1993]
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a)0°
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b) 45°
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c)90°
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d)0.14°
Explanation
1.47=sin30 / sinr1 sinr1=1/2×1.47 r1=19.88°1.46=sin30 / sinr2 sinr2=1/2×1.46 r2=20.02°r2 - r1=0.14°Answer: (d)
Q.13
Motion of wave front of wave has direction with wave motion [ Raj.PMT 1997]
0%
a) Parallel
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b) Perpendicular
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c)Opposite
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d)At angle of θ
Explanation
Answer: (a)
Q.14
If a plate of mica of thickness t and refractive index µ is introduced in the path of one of the coherent waves in interference experiment, the fringe pattern will shift by [ Raj.PMT 1997]
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a)d(µ-1) /D
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b) D(µ-1)t / d
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c)d/ [(µ-1)D]
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d) D(µ-1) / d
Explanation
Use lateral shift formula Answer:(b)
Q.15
In a Young's double slit experiment, the fringe width is found to be 0.4mm. If the whole apparatus is immersed in water of refractive index 4/3, without disturbing the geometrical arrangement, the new fringe width will be [ CBSE 1990]
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a) 0.30 mm
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b) 0.40mm
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c) 0.53mm
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d) 45 microns
Explanation
Answer: (a)
Q.16
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.17
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.18
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.19
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.20
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]
0%
a)0.5 mm
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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.21
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.22
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.23
Two slits separated by a distance of 1mm are illuminated with red light of wave length 6.5×10-7 m. The interference fringes are observed on screen placed 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to [ MPPMT 1995] [ MPPMT 1995]
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a) 0.65 mm
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b) 1.63 mm
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c) 3.25 mm
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d) 4.88 mm
Explanation
Position of third dark fringe Position of fifth bright fringe ∴ Difference in their positions=3.25 - 1.625=1.625=1.63 mmAnswer: (b)
Q.24
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
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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.25
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.26
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.27
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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b) 6.5×10-4 mm
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c) 1.3 micron
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d) 2.6×10-4 cm
Explanation
from the formula for minimum of diffraction a sinθ=nλ Answer: (c)
Q.28
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.29
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.30
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.31
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
0%
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.32
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.33
Light of wavelength 6328Å is incident on slit having a width of 0.2mm. the width of the central maximum, measured from minimum to minimum of diffraction pattern on a screen 9.0 m away will be about [ MPPMT 1987]
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a) 0.36°
0%
b) 0.18°
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c)0.72°
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d)0.09°
Explanation
Angular width of central maxima θ=2λ /d=θ=2×6328×10-10 / 0.2×10-3 θ=6328×10-6 θ=0.36°Answer: (a)
Q.34
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.35
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.36
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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a) λ=b2 / d
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b) λ=2b2 / d
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c)λ=3b2 /d
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d)λ=2b2 /3d
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.37
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.38
A beam of light of wavelength 600nm from a distance source falls on a single slit 1.00mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is [ IIT 1994]
0%
a)1.2cm
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b) 1.2 mm
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c)2.4 cm
0%
d)2.4 mm
Explanation
θ=2λ/d and θ=x/D X=D×2λ / dX=2×2×600×10-9 / 10-3 x=2400×10-6 m x=2.4mmAnswer: (d)
Q.39
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
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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.40
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]
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a) 0.8 mm
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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.41
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.42
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.43
The intensity ratio I1 / I2 of the two interfering sources in Young's experiment isThe ratio Imax / Imin is [ BHU 1995]
0%
a) 4:1
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b) 2:1
0%
c)3:1
0%
d)9:1
Explanation
Answer: (d)
Q.44
In Young's two slit interference experiment the distance between the slits is made 3 fold the fringewidth becomes [ CPMT 1989]
0%
a) 1/3 fold
0%
b) 3 fold
0%
c)1/9 fold
0%
d)9 fold
Explanation
Answer:(a)
Q.45
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
0%
c) 85
0%
d) 99
Explanation
Field view is same so 92×5898=n×5461 n=99 Answer: (d)
Q.46
Wave front means [ Raj.PMT 1997]
0%
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.47
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.48
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
0%
b) The intensity of light gradually increases to maximum and remains at maximum
0%
c)There is no change in intensity
0%
d)The intensity of light is twice maximum and twice zero
Explanation
Answer:(d)
Q.49
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.50
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
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d)Its velocity is very large
Explanation
Answer: (c)
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