Ray Optics
Q. If a concave lens has a focal length of -15 cm, what is its power?
A.
-6.67 D
B.
6.67 D
C.
-15 D
D.
15 D
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Solution
Power (P) = 1/f = 1/(-0.15) = -6.67 D.
Correct Answer: A — -6.67 D
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Q. If a concave mirror has a radius of curvature of 20 cm, what is its focal length?
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
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Solution
The focal length (f) of a concave mirror is half the radius of curvature (R). Therefore, f = R/2 = 20 cm / 2 = 10 cm.
Correct Answer: B — 10 cm
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Q. If a convex lens has a focal length of 15 cm, what is the power of the lens?
A.
+2.5 D
B.
+5 D
C.
+10 D
D.
+15 D
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Solution
Power (P) is given by P = 1/f (in meters). Thus, P = 1/0.15 = +6.67 D.
Correct Answer: B — +5 D
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Q. If a lens has a focal length of -20 cm, what type of lens is it?
A.
Convex lens
B.
Concave lens
C.
Biconvex lens
D.
Biconcave lens
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Solution
A negative focal length indicates that the lens is a concave lens.
Correct Answer: B — Concave lens
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Q. If a ray of light passes through the optical center of a lens, what happens to the ray?
A.
It bends towards the normal.
B.
It bends away from the normal.
C.
It continues in a straight line.
D.
It converges to a point.
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Solution
A ray of light passing through the optical center of a lens continues in a straight line without bending.
Correct Answer: C — It continues in a straight line.
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Q. If the angle of incidence is equal to the angle of refraction, what can be said about the two media?
A.
They are the same medium.
B.
They have the same refractive index.
C.
The light is traveling in a vacuum.
D.
The light is not refracted.
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Solution
When the angle of incidence equals the angle of refraction, it indicates that the light is passing from one medium to another of the same optical density, hence they are the same medium.
Correct Answer: A — They are the same medium.
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Q. If the angle of incidence is equal to the angle of refraction, what is the medium?
A.
Vacuum
B.
Air
C.
Glass
D.
Optically denser medium
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Solution
According to Snell's law, if the angle of incidence equals the angle of refraction, the light is traveling in the same medium, which can be vacuum or air.
Correct Answer: A — Vacuum
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Q. If the focal length of a concave lens is -10 cm, what is the nature of the image formed when an object is placed at 15 cm?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
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Solution
Concave lenses always produce virtual and erect images regardless of the object distance.
Correct Answer: B — Virtual and erect
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Q. If the focal length of a concave mirror is 10 cm, what is the radius of curvature?
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
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Solution
The radius of curvature (R) is twice the focal length (f). Therefore, R = 2f = 2 * 10 cm = 20 cm.
Correct Answer: B — 10 cm
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Q. If the focal length of a concave mirror is 20 cm, what is the radius of curvature?
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
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Solution
The radius of curvature (R) is twice the focal length (f). Therefore, R = 2f = 2 * 20 cm = 40 cm.
Correct Answer: B — 20 cm
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Q. If the focal length of a lens is 20 cm, what is the power of the lens?
A.
+5 D
B.
+10 D
C.
-5 D
D.
-10 D
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Solution
Power (P) = 1/f (in meters). Here, f = 0.2 m, so P = 1/0.2 = +5 D.
Correct Answer: B — +10 D
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Q. If the focal length of a lens is 25 cm, what is the power of the lens?
A.
+2 D
B.
+4 D
C.
+5 D
D.
+10 D
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Solution
Power (P) is given by P = 1/f (in meters). Thus, P = 1/0.25 m = +4 D.
Correct Answer: C — +5 D
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Q. If the radius of curvature of a concave mirror is 40 cm, what is its focal length?
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
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Solution
The focal length f of a mirror is given by f = R/2. Thus, f = 40 cm / 2 = 20 cm.
Correct Answer: B — 20 cm
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Q. If the radius of curvature of a spherical mirror is 40 cm, what is its focal length?
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
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Solution
The focal length (f) of a spherical mirror is given by f = R/2. Here, R = 40 cm, so f = 40/2 = 20 cm.
Correct Answer: B — 20 cm
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Q. If the refractive index of a medium is 1.5, what is the speed of light in that medium?
A.
2 x 10^8 m/s
B.
3 x 10^8 m/s
C.
1.5 x 10^8 m/s
D.
4.5 x 10^8 m/s
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Solution
The speed of light in a medium is given by v = c/n. Here, c = 3 x 10^8 m/s and n = 1.5, so v = 3 x 10^8 / 1.5 = 2 x 10^8 m/s.
Correct Answer: A — 2 x 10^8 m/s
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Q. In a thin lens, if the object distance is 15 cm and the image distance is 10 cm, what is the magnification?
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Solution
Magnification (m) is given by m = -v/u. Here, m = -10/15 = -2/3, which means the magnification is 1.5 in absolute value.
Correct Answer: B — 1.5
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Q. In a total internal reflection, what is the minimum angle of incidence for light traveling from water to air?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
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Solution
The critical angle for water to air is approximately 48.6 degrees. Therefore, the minimum angle of incidence is 60 degrees.
Correct Answer: C — 60 degrees
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Q. What happens to the focal length of a lens when it is immersed in a medium with a higher refractive index than the lens material?
A.
Focal length increases
B.
Focal length decreases
C.
Focal length remains the same
D.
Focal length becomes infinite
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Solution
When a lens is immersed in a medium with a higher refractive index, its effective focal length decreases due to the reduced refractive power.
Correct Answer: B — Focal length decreases
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Q. What happens to the image formed by a concave lens when the object is placed at infinity?
A.
Real and inverted
B.
Virtual and upright
C.
Real and upright
D.
No image formed
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Solution
When an object is placed at infinity, a concave lens forms a virtual image at its focal point, which is upright.
Correct Answer: B — Virtual and upright
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Q. What happens to the image formed by a concave mirror when the object is placed at the center of curvature?
A.
The image is virtual and upright.
B.
The image is real and inverted.
C.
The image is real and upright.
D.
The image is virtual and inverted.
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Solution
When the object is at the center of curvature, the image formed is real, inverted, and of the same size.
Correct Answer: B — The image is real and inverted.
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Q. What happens to the image formed by a concave mirror when the object is placed between the focal point and the mirror?
A.
The image is real and inverted.
B.
The image is virtual and upright.
C.
The image is real and upright.
D.
No image is formed.
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Solution
When the object is placed between the focal point and the mirror, the image formed is virtual and upright.
Correct Answer: B — The image is virtual and upright.
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Q. What happens to the image when the object is moved closer to a convex lens than its focal length?
A.
Image disappears
B.
Image becomes real
C.
Image becomes virtual
D.
Image becomes inverted
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Solution
When the object is within the focal length of a convex lens, the image formed is virtual.
Correct Answer: C — Image becomes virtual
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Q. What happens to the speed of light as it passes from air into a denser medium like glass?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
The speed of light decreases when it passes from a less dense medium (air) to a denser medium (glass) due to the change in refractive index.
Correct Answer: B — It decreases
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Q. What is the critical angle for total internal reflection from glass to air if the refractive index of glass is 1.5?
A.
30 degrees
B.
41.8 degrees
C.
48.6 degrees
D.
60 degrees
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Solution
Critical angle (C) is given by sin(C) = n2/n1. Thus, sin(C) = 1/1.5, C = sin^(-1)(0.6667) ≈ 41.8 degrees.
Correct Answer: B — 41.8 degrees
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Q. What is the critical angle for total internal reflection from water (n = 1.33) to air (n = 1)?
A.
48.6 degrees
B.
53.1 degrees
C.
60 degrees
D.
90 degrees
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Solution
The critical angle (θc) can be calculated using sin(θc) = n2/n1. Thus, sin(θc) = 1/1.33, giving θc ≈ 53.1 degrees.
Correct Answer: B — 53.1 degrees
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Q. What is the critical angle for total internal reflection if the refractive index of the medium is 1.5?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
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Solution
The critical angle (θc) can be calculated using sin(θc) = 1/n. Here, n = 1.5, so θc = sin^(-1)(1/1.5) ≈ 41.81 degrees, which is approximately 42 degrees.
Correct Answer: C — 60 degrees
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Q. What is the critical angle for total internal reflection when light travels from glass (n = 1.5) to air (n = 1)?
A.
30 degrees
B.
41.8 degrees
C.
48.6 degrees
D.
60 degrees
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Solution
The critical angle θc is given by sin(θc) = n2/n1. Thus, θc = sin^(-1)(1/1.5) ≈ 41.8 degrees.
Correct Answer: B — 41.8 degrees
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Q. What is the critical angle for total internal reflection when light travels from water (n = 1.33) to air (n = 1)?
A.
48.6 degrees
B.
53.1 degrees
C.
60 degrees
D.
90 degrees
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Solution
The critical angle (θc) can be calculated using sin(θc) = n2/n1. Here, n1 = 1.33 (water) and n2 = 1 (air). Thus, sin(θc) = 1/1.33, giving θc ≈ 53.1 degrees.
Correct Answer: B — 53.1 degrees
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Q. What is the focal length of a concave mirror if an object is placed at a distance of 30 cm from the mirror and the image is formed at a distance of 15 cm from the mirror?
A.
10 cm
B.
15 cm
C.
20 cm
D.
25 cm
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Solution
Using the mirror formula, 1/f = 1/v + 1/u. Here, v = -15 cm (real image) and u = -30 cm (object distance). Thus, 1/f = 1/(-15) + 1/(-30) = -1/10, so f = -10 cm.
Correct Answer: A — 10 cm
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Q. What is the focal length of a concave mirror if an object placed 30 cm in front of it produces an image at 10 cm in front of the mirror?
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
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Solution
Using the mirror formula 1/f = 1/v + 1/u, where u = -30 cm and v = -10 cm, we find f = 15 cm.
Correct Answer: C — 15 cm
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