Physics numerical in English:- chapter- light reflection and refraction
Q.1. If the focal length of the lens is 10 cm, then what will be the power of the lens?
A) 10D
B) 0.1D
C) 5D
D) 20D
ANSWER= A) 10D.
Explain:- 1. If the focal length is given in cm, then the power of the lens will be :-
100/f
2. If the focus distance is given in m, then the power of the lens will be :-
1/f
Here it is given in cm.
Hence the power of the lens = 100/f
= 100/10
= 10D
Q.2. What will be the focal length of the lens whose power is +2 D :-
A) 10cm
B) 5cm
C) 50 cm
D) 20cm
ANSWER= C) 50cm.
Explain:- Here the focus distance is given in cm.
Hence, the power of the lens = 100/f
2 = 100/f
f =100/2
= 50 cm
Q.3. The magnifying power of an astronomical telescope is 10. If the focal length of the objective lens is 1 metre, then what will be the focal length of the eyepiece lens?
A) 0.1m
B) 1m
C) 1.5m
D) 2.5m
ANSWER= A) 0.1m
Explain:- here fo = 1 m fe = ? M = 10
M = fo/fe
M= 1/fe
fe = 1/10
= 0.1 m
Q. 4. A concave mirror is made by cutting a spherical lens of radius 10 cm. The radius of curvature (R) of this mirror will be :-
A) 20 cm
B) 10 cm
C) 5 cm
D) 15 cm
ANSWER= B) 10 cm
Explanation:- The radius of the spherical lens, which is cut into a mirror, is the radius of curvature of the mirror.
Q.5. If the focal length of a convex mirror is 15 cm, then what is the distance between the pole o of the mirror and the center of curvature C?
A) 20 cm
B) 10 cm
C) 30 cm
D) 15 cm
ANSWER= C) 30 cm
Explain:- The distance between the pole and the center of curvature, the radius of curvature, is R.
R = 2f
= 2×15
= 30 cm
Q.6. A concave mirror is made by cutting a spherical lens of diameter 40 cm. If the object is placed at infinity, what will be the image distance?
A) 12 cm
B) 10 cm
C) 30 cm
D) 15 cm
ANSWER= B) 10 cm
Explain:-
Radius of spherical lens of diameter 40cm = 20 cm
Therefore, radius of curvature of a concave mirror R = 20cm
Focus distance f = R/2
= 20/2
=10cm
When the object is placed at infinity in a concave mirror, the focus is formed at the point. Hence image distance = f
=10cm
Q.7. If the radius of curvature of a mirror is 60 cm, what is its focal length?
A) 12 cm
B) 10 cm
C) 30 cm
D) 15 cm
ANSWER= C) 30 cm
Explain:- f = R/2
=60/2
= 30cm
Q.8. An object is placed at a distance of 20 cm in front of a convex mirror of focal length 20 cm. Where will the image of the object be formed?
A) 12 cm
B) 15 cm
C) 30 cm
D) 10 cm
ANSWER= D) 10 cm
Explain:- In a convex mirror,
Focus distance = f = 20 cm Object distance = -u = -20 cm Image distance = +v
1/f= 1/v+1/u
1/v= 1/f-1/u
1/v = {1/20-(1/-20)}
= (1/20+1/20)
=2/20
=1/10
v = 10 cm
Q.9. A ray of light falls on a plane mirror at an angle of incidence of 60°. The angle between the reflected light ray and the surface of the mirror will be :-
A) 30°
B) 60°
C) 90°
D) 45°
ANSWER= A) 30°
Explain:- Angle between normal and mirror surface = 90°
By the law of reflection of light,
The angle between the normal and the reflected ray = 60°
Therefore, angle between the reflected ray and mirror surface = 90°-60°
= 30°
Q.10. Two plane mirrors are placed making an angle of 60° to each other. If an object is placed between the two mirrors, how many total images will be formed?
A) 3
B) 6
C) 9
D) 5
ANSWER= D) 5
Explain:- {[ Formula :- n ={ (360°/α)-1}]
n = (360/60)-1
= 6-1
=5
Q.11. An object of height 2 mm is placed at a distance of 20 cm in front of a convex mirror. If the radius of curvature of the mirror is 40 cm, find the height of the image.
A) 3mm
B) 1mm
C) 4mm
D) 5mm
ANSWER= B) 1mm
Explain:- Magnification m = I/O = f/(f-u)
So I/.20= [{20/{20-(-20)}]
I = .40/40
= .10cm
= 1mm
{ Note :- Formula:- m= f/(f-u) 10mm= 1cm 1mm= 1/10=.10cm}
Q.12. The refractive index of water is 1.33. What will be the speed of light in water?
A) 3.33 × 10⁸ m/s
B) 1.12 × 10⁸ m/s
C) 5.76 × 10⁸ m/s
D) 2.25 × 10⁸ m/s
ANSWER= D) 2.25 × 10⁸ m/s
Explain:-
Refractive index of water = speed of light in air / speed of light in water
1.33 = 3 × 10⁸m/s/speed of light in water
Velocity of light in water =( 3×10⁸)÷1.33 m/s
= 2.25× 10⁸ m/s
Q.13. The refractive index of glass is 3/2. What will be the speed of light in the glass?
A) 3 × 10¹⁰ cm/s
B) 2 × 10¹⁰ cm/s
C) 7 × 10¹⁰ cm/s
D) 1.5 × 10¹⁰ cm/s
ANSWER= B) 2 × 10¹⁰ cm/s
Explain:-
Refractive index of glass = speed of light in air / speed of light in glass
Therefore, velocity of light in glass = velocity of light in air / refractive index of glass
= 3× 10⁸ /(3 /2 )m/s
=2 ×10⁸ m/s
=2 ×10¹⁰ cm/s
Q.14. The refractive index of water is 4/3 and that of glass is 3/2. What will be the refractive index of glass relative to water?
A) 5/4
B) 1/2
C) 9/8
D) 7/5
ANSWER= C) 9/8
Explain:-
Q.15. The refractive index of a medium is 2. If light travels through that medium into air, then what will be the value of the critical angle?
A) 30°
B) 45°
C) 60°
D) 15°
ANSWER= B) 45°
Explain:- sinC = 1/μ
=1/√2
=sin45°
So C = 45°
Q.16. The velocity of light is 3 × 10⁸ m/s in air and 2 × 10⁸ m/s in water. Then what will be the value of critical angle?
A) Sin-¹(2/3)
B) Sin-¹(3/2)
C) Sin-¹(4/3)
D) Sin-¹(2/5)
ANSWER= A) Sin-¹(2/3)
Explanation:- The critical angle occurs only when light passes from a denser medium to a rarer medium. Therefore light travels from water to air.
Refractive index of water μ=3 ×10⁸/2×10⁸
= 3/2
Now, sinC = 1/μ
= 1/(3/2)
=2/3
C= sin-1(3/2)
Q.17. An object is placed in the axis of a convex lens. The image is formed at a distance of 80 cm from the object. If the magnifying power of the lens is 3 then find the focal length of the lens is:-
A) 15 cm
B) 25cm
C) 10cm
D) 24cm
ANSWER= A) 15 cm
Explain:- The distance between the object and the image is 80 cm . If the image distance is v, then the object distance will be (80 -v).
Magnifying power of the lens m = v/u
3 = v/(80 -v )
240 -3v =v
v +3v =240
4v = 240
v =240/4
=60 cm
so u = 80 -60
= 20 cm
Now 1/f = 1/v -1/u
=1/60 -(1/-20 )
= 1/60 +1/20
=(3+1 )/60
=4/60
=1/15
f = 15 cm
Note :- For calculation f and v in convex lens are taken in + while u is taken in -.
Q.18. An object of height 5cm is placed at a distance of 2m in front of a concave mirror. If the radius of curvature of a concave mirror is 40 cm, find the height of the image.
A) 0.15cm
B) 0.25cm
C) 0.55cm
D) 0.40cm
ANSWER= C) 0.55 cm
Explain:- Object distance = 2 m = 200 cm Focus distance f = 20 cm (Half of radius of curvature) Image distance = v
From the mirror formula, 1 /f = 1 /v +1 /u
1/-20 = 1/v +(1/-200 )
1/v = 1/200 -1/20
= (1 -10)/200 ,
= -9/200
v = 200/9 cm
Now, (height of the image/height of the object)= -v/u
Image height/5 = -(200/9)/200
Image height = -5/9
= -0.55
Hence height of image = 0.55 cm
Note : In a concave mirror only the height of the object is in +.
Q.19. The minimum angle of deviation of an equilateral glass prism is 30°. What is the refractive index of a prism?
A) 3.5
B) 2
C) 4.5
D) 1/2
ANSWER= B) √2
Explain:- All angles in an equilateral prism are of 60°.
Therefore, prism angle A =60°
Minimum deviation angle = 30°
Refractive index μ = sin{(A+δ)/2}÷sinA/2
= sin{(60+30)/2}÷sin(60/2)
= sin45°÷sin30°
=(1/√2)÷1/2
= 2/√2
= (2×√2)÷(√2×√2)
= 2√2/2
= 2
Q. 20. The value of angle of minimum deviation of an equilateral prism is 60°. Find the refractive index of this prism.
A) 3.57
B) 1.21
C) 1.114
D) 1.732
ANSWER= D) 1.732
Explain:- refractive index μ = sin{(A+δ)/2}÷sinA/2
=sin{(60+60)/2}÷ sin(60/2)
= sin60° sin30°
= (√3/2)÷(1/2)
= 3
= 1.732
Q.21. An object is placed at the center of curvature in front of a concave mirror. If the focal length of the concave mirror is 10 cm, find the image distance.
A) 30cm
B) 20 cm
C) 15cm
D) 25cm
ANSWER= B) 20 cm
Explanation:- If the object is placed at the center of curvature in a concave mirror, the image is formed only at the center of curvature.
So reflection distance = R
= 2f
= 2×10
= 20cm
Q.22. The object is placed in front of a convex lens of focal length 15cm. If the image is formed at a distance of 20 cm, then what is the distance of the object?
A) 60cm
B) 80 cm
C) 65cm
D) 45cm
ANSWER= A) 60 cm
Explanation:- In a convex lens, the focus distance and image distance are in + while the object is in distance -.
So f = +15 cm v = +20 cm object distance = -u
From the lens formula, 1 /f = 1 /v -1 /u
1 /15 = 1 /20 -(1 /-u )
1 /15 =1 /20 +1 /u
1 /u = 1 /15 -1 /20
= (4 -3) /60
= 1 /60
u = 60 cm
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