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Physics NEET MCQ
Quiz 10
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Q.1
A concave mirror has focal length 20cm. The distance between the two position of object for which the image size i doubled the object size is
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a) 20 cm
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b) 40 cm
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c)30 cm
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d)60 cm
Explanation
For real image u=-u1 then form the formula for magnification v==-2u1 f=-20 cm Substituting the data in mirror formula we get u1=30 cm For virtual image u=-u2 , and v=2u2 f=-20 cm Substituting the data in mirror formula we get u2=10 cm Distance between two position of the object=u1 -u2=20 cmAnswer: (a)
Q.2
One side of the glass slab is silvered as shown. A ray of light is incident on the other side at an angle of incidence i=45°. Refractive index of glass is given as 1.The deviation of the ray of light from its initial path when it comes out of slab is
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a) 90°
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b) 180°
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c)120°
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d)45°
Explanation
Answer:(a)
Q.3
A thin prism with angle 4° and refractive index 1.5 is placed inside a transparent tube with water ( refractive index=5/4) as shown. The deviation of light due t prism will be
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a) 0.8° upward
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b) 0.8° downward
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c) 0.67° upward
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d) 0.37° downward
Explanation
Let µ1 be the refractive index of prism and µ2 be the refractive index of water Answer: (b)
Q.4
A parallel narrow beam of light is incident on the surface of a transparent hemisphere of radius R and refractive index µ=1.5 as shown in figure. The position of the image formed by the refraction at the spherical surface only is
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a)R/2
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b) 3R
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c)R/3
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d)2R
Explanation
Object is at infinity, and image is in glass use formula form refraction at curved surface Answer: (b)
Q.5
A man approaches a vertical plane mirror at speed of 2m/s. Then the rate at which he approaches his image is
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a) 2 m/s
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b) 4m/s
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c)1 m/s
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d)zero
Explanation
Answer: (b)
Q.6
A ray of light falls on the surface of a spherical glass paper-weight making an angle α with normal and is refracted in the medium at an angle β. The angle of deviation of emergent ray from the direction of the incident ray is
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a) (α - β)
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b) 2(α - β)
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c)(α - β) / 2
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d)(α + β)
Explanation
from figure δ=2(α - β) Answer:(b)
Q.7
If white light is used in a Young's double slit experiment
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a) bright white fringe is not formed at the centre of the screen
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b) fringes of different colurs are not observed clearly only in the first order
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c) the first-order violet fringe is closer to the central fringe than the first-order red fringe
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d) the first order red fringe is closer to the central fringe than the first-order violet fringe
Explanation
With white light only central fringe is white As λviolet < λred Position of fringe X=Dnλ / d Xviolet < Xred Condition for constructive interference is satisfied by violet first and the red. The violet fringe will be next to central fringe Answer: (c)
Q.8
A thin lens of focal length f produces an upright image of same size as the object. What is the distance of the object from the optical centre of the lens?
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a)2f
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b) zero
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c)3f/2
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d)infinity
Explanation
v=mu given m=1 v=u is possible for convex lens only if v=u=0Answer: (b)
Q.9
Two thin lenses, one concave and the other is convex are placed in contact with each other. If their powers are in the ratio 2/3 and the effective focal length of the combination is 30cm, then the individual focal lengths are
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a) -75 cm, +50 cm
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b) -15 cm, 10cm
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c)+75cm, -50cm
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d)75cm, 50cm
Explanation
on simplification of above equations focal length of concave lens=-15cm and focal length of convex lens=+10cmAnswer: (b)
Q.10
Two identical sources P and Q emit waves in same phase and same wavelength. Spacing between P and Q is 3λ. The maximum distance from P along the x-axis at which a minimum intensity occurs is given by
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a) 6.58λ
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b) 2.25λ
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c)8.75λ
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d)055λ
Explanation
On the horizontal, the path difference is decreasing on increasing X . so at the maximum distance from point P where minima occurs, path difference is λ/2 from geometry of figure Answer:(c)
Q.11
Interference pattern is obtained by using light having two wave lengths 400nm and 560nm. The light falls normally on two narrow slits. The distance between the successive regions of "Total Darkness' is
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a) 4 mm
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b) 5.6 mm
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c) 14mm
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d) 28mm
Explanation
region between Total Darkness is bright fringe At a point constructive interference of both should overlap by using formula Answer: (d)
Q.12
The refractive index of material varies with wavelength according to the Cauch's relation is
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a) µ=A λ + B
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b) µ=A + B / λ2
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c)µ=A × B / λ
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d)µ=A2λ + B
Explanation
Answer: (b)
Q.13
Indicate the correct statement in the following
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a)The dispersive power depends upon the angle of prism
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b) The angular dispersion depends upon the angle of the prism
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c)the angular dispersion does not depend upon the dispersive power
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d)the dispersive power in vacuum is one
Explanation
Answer: (b)
Q.14
A ray of light falls normally on a refracting face of a prism of refractive index 1.What is the angle of prism if the ray just fails to emerge from the prism?
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a)30°
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d)none of these
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b) sin⁻¹ (1/3)
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c)tan⁻¹ (3/2)
Explanation
For total internal reflection at face AB.A=r1 + r2 r1=0 on Ac ∴ r2=ABut r2 > C (Critical angle) sinA > 1 /µsinA > (2/3) Answer: (d)
Q.15
Light is incident normally on face AB of a prism as shown in figure. A liquid of refractive index µ is placed on the face AC of the prism. The prism is made of glass of refractive index 3/The limits of µ for which total internal reflection takes place on face AC is
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a) µ > √3 /2
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b) µ < 3√3 /4
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c)µ > √3
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d)µ < √3 /2
Explanation
Angle of incidence on face AC=60°For total internal reflection angle of incidence on face AC sinθ > (3/2)/µ sin60 > 2µ/3 √3 / 2 > 2µ/33√3 / 4 > µµ < 3√3 / 4Answer: (b)
Q.16
A ray of light passing through a prism of refracting angle 60° has to deviate by at least 30°. Then refractive index of prism should be
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a) ≤ √2
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b) ≥ √2
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c)≥ √3
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d)≤ √3
Explanation
Use formula On simplification we get µ=√2, At µ=√2, minimum deviation is 30°, for δ ≥ 30°, µ ≥ √2 Answer:(b)
Q.17
An object is kept at a distance of 16 cm from a think lens and the image formed is real. If the object is kept at distance of 6cm from the same lens, the image formed is virtual. If the size of the images formed are equal, the focal length of the lens will be
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a) 15 cm
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b) 17 cm
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c) 21 cm
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d) 11 cm
Explanation
Real image is formed only by convex lens u=-16 cm if n is magnification then v=16n applying lens equation we get second case u=-6 cm v=-6n ( as magnification is same in both cases) adding (1) and (2) we get 2=22/f or f=11cmAnswer: (d)
Q.18
Young's double slit experiment is made in liquid. the 10th bright fringe lies in liquid where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately.
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a)1.8
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b) 1.5
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c)1.3
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d)1.6
Explanation
Position of fringe in both case is same use formula for dark and bright fringeLet refractive index of liquid be Let wave length of light in liquid be λ'=λ/µ Answer: (a)
Q.19
In Young's double slit experiment how many maximas can be obtained on a screen ( including central maximum) on both sides of the central fringe, if λ=2000Å and d=7000Å. Given that perpendicular distance of screen from the mid point of two slits is 3.5 cm
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a) 12
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b) 7
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c)18
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d)4
Explanation
For maximum intensity on the screendsinθ=nλ sinθ=nλ/d Since can have maximum value=1nλ/d=1 n=d/ λ=7000 / 2000=3.5 thus n=3 that is 3 bright fringe up three down total fringes seven including central maximaAnswer: (b)
Q.20
A plastic hemisphere has a radius of curvature of 8cm and an index of refraction of 1.On the axis half way between the plane surface and the spherical one ( 4 cm from each) is small object O. The distance between the two images when viewed along the axis from the two sides of the hemisphere is approximately
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a) 1.0 cm
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b) 1.5 cm
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c)3.75 cm
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d)2.5 cm
Explanation
Distance of image from plane surface is distance apparent=distance actual / µ x1=4/1.5=2.5 cm Distance of image from curved surface ( use formula for refraction at curved surface) negative sign indicates image is on object side Distance between images=( 8 - 2.5 -3)=2.5 cm Answer:(d)
Q.21
A ray is incident on a medium consisting of two boundaries, one plane and other curved as shown in figure . The plane surface makes an angle of 60° with horizontal and curved surface has radius of curvature 0.4m. The refractive indices of the medium and its environment are shown in the figure. If after refraction at both the surfaces the ray meets the principle axis at P, find OP
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a) 2.056 m
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b) 5.056 m
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c) 6.056 m
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d) 4.056 m
Explanation
Applying Snell's law at face AB 1×sin45=√2 ×sinr ∴ r=30° Refracted ray will be parallel to the face AD as shown in figure Now refraction at spherical surface u=∞ ( as incident ray is parallel to principal axis ) Answer: (c)
Q.22
A ray of light is incident on a glass sphere of refractive index 3/What should be the angle of incidence so that the ray which enters the sphere comes out tangentially from the surface of the sphere
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a)tan⁻¹(2/3)
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b) sin⁻¹(2/3)
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c)90°
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d)cos⁻¹(1/3)
Explanation
As shown in figure ∠ABO=∠OAB=θApplying snells law at A sini=sinθ(3/2) --eq(1)Applying Senll's law at B (3/2)sinθ=sin90 sinθ=2/3 --eq(2)from equation (1) and (2) sini=(2/3)(3/2)=1 thus i=90°Answer: (c)
Q.23
A plane mirror is made of glass slab (µg=1.5), 2.5 cm thick and silvered on the back. A point object is placed 5cm in front of the unsilvered face of the mirror. What will be the position of final image?
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a) 12 cm from unsilvered face
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b) 14.6 cm from unsilvered face
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c)5.67 cm from unsilvered face
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d)8.33 cm from unsilvered face
Explanation
As shown in figure I2 : Image formed due to reflection from DEFI3 : Image form due to refraction from ABCObject distance for face DEF Thus image I2=7.5 +2.5=10cm behind face DFF This image will act as object , distance from face ABC=10+2.5=12.5 image is behind the unsilvered faceAnswer: (d)
Q.24
The maximum intensity in Young's double slit experiment is Io. Distance between the slits is d=5λ, where λ is the wave length of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on screen at a distance D=10d?
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a) Io /2
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b) (3/4)Io
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c)Io
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d)Io/4
Explanation
Position of point onscreen x then path difference=xd/D Here point is at distance x=d/2 to make it infrant of one of the slit given d=5λ and D=10d=50λCorresponding phase difference φ=π/2From the formula for intensity when Io is maximum intensity Answer:(a)
Q.25
In Young's double slit experiment d/D=10-At a point P on the screen the resulting intensity is equal to the intensity due to individual slit Io. Then the distance of point P from the central maxima is ( λ=6000Å)
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a) 2 mm
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b) 1 mm
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c) 0.5 mm
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d) 4 mm
Explanation
If Io is intensity due to individual slit then From the formula for Intensity at point, Path difference for 2π /3=Λ/3 But Path difference=xd/D Answer: (a)
Q.26
If the critical angle for the medium of prism is C and the angle of the prism is A, then there will be no emergent ray when
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a) A < 2C
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b) A=2C
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c)A > 2C
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d)A >=< 2C
Explanation
Answer:(c)
Q.27
Longitudinal chromatic aberration is given by
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a) ω + f
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b) ω × f
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c)ω / f
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d)ω f2
Explanation
Answer: (b)
Q.28
A point object is placed at a distance of 25cm from a convex lens of focal length 20cm. If a glass slab of thickness to and refractive index 1.5 is inserted between the lens and the object the image is formed at infinity, the thickness t is
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a)10 cm
0%
b) 5 cm
0%
c)20 cm
0%
d)15 cm
Explanation
Image will be form at infinity if object is placed at focus of the lens. That is 20 cm from the lensThus shift=25 -20=5Formula for shift=[ 1- (1/µ)]tf=[ 1 - (1/1.5)]t t=15 cmAnswer: (d)
Q.29
lens of glass (µg=1.5) of focal length 10 cm is silvered on one side. It will behave like a
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a) a) concave mirror of focal length 10 cm
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b) convex mirror of focal length 5.0 cm
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c)concave mirror of focal length 2.5 cm
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d)convex mirror of focal length 20 cm
Explanation
For equiconvex lens |R1|=|R2|=f=10 cm Therefore, the system will behave like concave mirror of focal length 2.5 cmAnswer: (c)
Q.30
A monochromatic beam of light falls on Young's double slit Experiment apparatus at some angle as shown in figure.A thin sheet of glass is inserted in front of the lower slit SThe central bright fringe will be obtained
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a) at O
0%
b) above O
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c) below O
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d) anywhere depending on thickness of plate angle and refractive index of glass
Explanation
If path difference dsinθ=(µ - 1) t, central fringe is obtained at O If dsinθ > (µ - 1) t, central fringe is obtained above O If dsinθ < (µ - 1) t, central fringe is obtained below OThus fringe position depends on thickness of plate angle and refractive index of glass Answer: (d)
Q.31
A plano convex lens (µg=3.2) of radius of curvature R=10 cm is placed at a distance 'b' from a concave lens of focal length 20 cm. What should be the distance 'a' of a point object O from the plano-convex lance so that the position of final image is independent of 'b'?
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a)40 cm
0%
b) 60 cm
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c)30 cm
0%
d)20 cm
Explanation
Focal length of plano-convex lens is If object O is placed at a distance of 20 cm from the plano-convex lens rays become parallel and final image is formed at second focus of 20 cm from concave lens which is independent of bAnswer: (d)
Q.32
One of the refracting surface of a prism of angle 30° is silvered. A ray of light incident at an angle of 60° retraces its path. The refractive index of the material of the prism is
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a) √2
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b) √3
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c)3/2
0%
d)2
Explanation
From figure r2=0 r1=A=30 and i1=60° µ=sini1 / sinr1=√3Answer: (b)
Q.33
A ray of light falls on a transparent sphere with centre at C as shown in figure. The ray emerges from the sphere parallel to line AB. The refractive index of sphere is
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a) √2
0%
b) √3
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c)3/2
0%
d)1/2
Explanation
Figure deviation by sphere is δ=2 ( i-r) HEre δ=60° 60=2(i-r) i- r=30 r=60 - 30=30° Now µ=sini/ sinr µ=sin60/sin30=√3 Answer:(b)
Q.34
Two identical glass (µg=3/2) Reunion lenses of focal length f are kept in contact. The space between the two lenses is filled with water (µw=4/3). The focal length of the combination is
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a) f
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b) f/2
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c) 4f/3
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d) 3f/4
Explanation
Let R be the radius of curvature of each surface Then For ware lensNow by using formula for combination of lenses Answer: (d)
Q.35
Angle of minimum deviation is equal to the angle of prism A of an equilateral glass prism. The angle of incidence at which minimum deviation will be obtained is
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a)60°
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b) 30°
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c)45°
0%
d)sin⁻¹ (2/3)
Explanation
A=δm=60°At minimum deviation i=[( A+δm) / 2]=60°Answer: (a)
Q.36
Refraction takes place at a concave spherical boundary separating glass air medium. For the image to be real, the object distance ( µg=3/2)
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a) should be greater than three times the radius of curvature of the refracting surface
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b) should be greater than two times the radius of curvature of the refracting surface
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c)should be greater than the radius of curvature of the refracting surface
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d)is independent of the radius of curvature of the refracting surface
Explanation
Object is in denser medium . Applying formula for refraction at curved sourceIn given problem for image to be real v should be positive which possible only if 1/ 2R > (3/2u)u > 3R Answer: (a)
Q.37
Mark the correct option
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a) If the incident rays are converging, we have real object
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b) If the final rays are converging, we have a real image
0%
c)The image of virtual object is called a virtual image
0%
d)If the image is virtual, the corresponding object is called a virtual object
Explanation
Answer:(b)
Q.38
A point source of light is placed in front of a plane mirror
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a) All the reflected rays meet at a point when produced backward
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b) Only the reflected rays close to the normal meet at a point when produced backward
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c) Only the reflected rays making a small angle with the mirror, meet at a point when produced backward
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d) Light of different colours make different images
Explanation
Answer: (a)
Q.39
On reflection from a plane surface, the following gets changed
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a)wavelength
0%
b) frequency
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c)speed
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d)amplitude
Explanation
Answer: (d)
Q.40
A plane mirror is approaching you at 10cm per second. You can see your image in it. At what speed will your image approach you
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a) 10 cm/s
0%
b) 5 cm/s
0%
c)20 cm/s
0%
d)15 cm/s
Explanation
Answer: (c)
Q.41
An object is approaching a plane mirror at 5 cms per second. A stationary observer sees the image. At what speed will the image approach the stationary observer
0%
a) 5 cms per second
0%
b) 20 cms per second
0%
c)10 cms per second
0%
d)15 cms per second
Explanation
Answer:(a)
Q.42
Which of the following letters do not suffers later inversion
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a) HGA
0%
b) HOX
0%
c) VET
0%
d) YUL
Explanation
Answer: (b)
Q.43
A small object is 10 cm in front of a plane mirror. A man stands 30 cm from the mirror behind the object and looks at the object's image. He should focus his eyes to see the image at a distance
0%
a)25 cm
0%
b) 35 cm
0%
c)45 cm
0%
d)40 cm
Explanation
Answer: (d)
Q.44
If you want to see your full image, then minimum size of the mirror
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a) Should be of your height
0%
b) Should be half of your height
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c)Should be twice of your height
0%
d)Depends upon distance from the mirror
Explanation
Answer: (b)
Q.45
In a concave mirror an object is placed at a distance x1 from the focus and the image is formed at a distance x2 from the focus. Then the focal length of the mirror is
0%
a)x1x2
0%
b)√(x1x2)
0%
c)(x₁+x₂) /2
0%
d)√(x₁/x₂)
Explanation
Answer:(b)
Q.46
If an object is placed unsymmetrical between two plane mirror, inclined at an angle of 72°, then the total number of images formed is
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a) 5
0%
b) 4
0%
c) 2
0%
d) infinity
Explanation
number images=360/θ Answer: (a)
Q.47
If the critical angle for total internal reflection from a medium to vacuum is 30° the velocity o light in the medium is
0%
a)3×108 m/s
0%
b) 1.5×108 m/s
0%
c)6×108 m/s
0%
d)√3×108 m/s
Explanation
use sinc=1/µand µ=c/vAnswer: (b)
Q.48
Light travels through a glass of thickness t and refractive index µ. If c is the velocity of light in vacuum, the time taken y light to travel through the plate is
0%
a) t/µc
0%
b) µtc
0%
c)µt/c
0%
d)tc/µ
Explanation
Answer: (c)
Q.49
If I1 and I2 be the size of the images respectively for two positions of the lens in the displacement method, then the size of the object is given by
0%
a)I1 I2
0%
b) √(I1 I2 )
0%
c) √(I₁ / I₂ )
0%
d)√(I₂ / I₁ )
Explanation
Answer:(b)
Q.50
Two thin lenses of focal length f1 and f2 are in contact and coaxial. Its power is same as the power of single lens given by
0%
a) (f₁ + f₂) /(f₁ × f₂)
0%
b) √(f₁ / f₂)
0%
c) √(f₂ / f₁)
0%
d) (f₁ + f₂) /2
Explanation
Answer: (a)
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