Engineering & Architecture Admissions
Q. A lens has a focal length of 40 cm. If an object is placed 80 cm from the lens, what is the image distance?
A.
40 cm
B.
60 cm
C.
80 cm
D.
100 cm
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Solution
Using the lens formula, we find the image distance to be 40 cm.
Correct Answer: A — 40 cm
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Q. A lens has a focal length of 50 cm. If an object is placed at 100 cm, what type of image is formed?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
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Solution
Since the object distance is greater than the focal length, a real and inverted image is formed.
Correct Answer: A — Real and inverted
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Q. A lens has a power of +2 diopters. What is its focal length?
A.
0.5 m
B.
1 m
C.
2 m
D.
3 m
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Solution
Power (P) is given by P = 1/f (in meters). Thus, f = 1/P = 1/2 = 0.5 m.
Correct Answer: B — 1 m
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Q. A lens has a power of +2.0 D. What is its focal length?
A.
50 cm
B.
25 cm
C.
20 cm
D.
10 cm
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Solution
Power (P) = 1/f (in meters), so f = 1/2.0 = 0.5 m = 50 cm.
Correct Answer: B — 25 cm
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Q. A lens has a power of +2.5 D. What is its focal length?
A.
40 cm
B.
25 cm
C.
50 cm
D.
20 cm
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Solution
Power (P) = 1/f (in meters). Therefore, f = 1/2.5 = 0.4 m = 40 cm.
Correct Answer: B — 25 cm
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Q. A lens has a power of +2.5 D. What is the focal length of the lens in meters?
A.
0.4 m
B.
0.5 m
C.
0.6 m
D.
0.7 m
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Solution
Power (P) = 1/f, so f = 1/P = 1/2.5 = 0.4 m.
Correct Answer: B — 0.5 m
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Q. A lens has a power of +5 diopters. What is its focal length?
A.
20 cm
B.
25 cm
C.
30 cm
D.
15 cm
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Solution
Power (P) = 1/f (in meters). Therefore, f = 1/P = 1/5 = 0.2 m = 20 cm.
Correct Answer: A — 20 cm
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Q. A lens has a power of -4 D. What is the type of lens?
A.
Convex
B.
Concave
C.
Bifocal
D.
Plano-convex
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Solution
A negative power indicates a concave lens.
Correct Answer: B — Concave
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Q. A lens has a power of -4 D. What type of lens is it?
A.
Convex lens
B.
Concave lens
C.
Bifocal lens
D.
Plano-convex lens
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Solution
A negative power indicates that the lens is a concave lens.
Correct Answer: B — Concave lens
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Q. A lens produces a magnification of 3. If the object height is 2 cm, what is the image height?
A.
4 cm
B.
6 cm
C.
3 cm
D.
2 cm
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Solution
Magnification (m) = h'/h, thus h' = m * h = 3 * 2 = 6 cm.
Correct Answer: A — 4 cm
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Q. A lens produces a virtual image at a distance of 12 cm when the object is placed at 8 cm. What type of lens is it?
A.
Convex
B.
Concave
C.
Biconvex
D.
Biconcave
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Solution
A virtual image is formed by a concave lens when the object is placed in front of it.
Correct Answer: B — Concave
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Q. A light bulb uses 60 W of power. How much energy does it consume in 1 hour?
A.
3600 J
B.
216000 J
C.
60000 J
D.
180000 J
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Solution
Energy consumed can be calculated using E = P * t. Here, P = 60 W and t = 1 hour = 3600 seconds. Thus, E = 60 W * 3600 s = 216000 J.
Correct Answer: B — 216000 J
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Q. A light bulb uses 60 W of power. How much energy does it consume in 2 hours?
A.
120 J
B.
7200 J
C.
432000 J
D.
360 J
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Solution
Energy consumed can be calculated using E = P * t. Here, P = 60 W and t = 2 hours = 7200 seconds. Thus, E = 60 W * 7200 s = 432000 J.
Correct Answer: C — 432000 J
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Q. A light bulb uses 60 W of power. How much energy does it consume in 5 hours?
A.
18000 J
B.
108000 J
C.
300000 J
D.
360000 J
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Solution
Energy consumed can be calculated using E = P * t. Here, E = 60 W * (5 * 3600 s) = 108000 J.
Correct Answer: B — 108000 J
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Q. A light bulb uses 60 Watts of power. How much energy does it consume in 2 hours?
A.
120 Joules
B.
7200 Joules
C.
432000 Joules
D.
360000 Joules
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Solution
Energy consumed = Power × Time = 60 W × (2 × 3600 s) = 432000 J.
Correct Answer: C — 432000 Joules
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Q. A light ray enters a glass prism with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction?
A.
15 degrees
B.
20 degrees
C.
25 degrees
D.
30 degrees
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Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5, i = 30 degrees. Solving gives r = 19.2 degrees.
Correct Answer: B — 20 degrees
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Q. A light ray in glass (n=1.5) strikes the boundary with air at an angle of 30°. Will it undergo total internal reflection?
A.
Yes
B.
No
C.
Depends on the angle
D.
Not enough information
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Solution
Since the angle of incidence is less than the critical angle (θc = sin⁻¹(1/1.5) ≈ 41.8°), it will not undergo total internal reflection.
Correct Answer: B — No
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Q. A light ray passes from air into glass with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction?
A.
18.4 degrees
B.
20 degrees
C.
22 degrees
D.
25 degrees
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Solution
Using Snell's law, n1 * sin(theta1) = n2 * sin(theta2). Here, n1 = 1 (air), n2 = 1.5 (glass), and theta1 = 30 degrees. Thus, sin(theta2) = (1 * sin(30))/1.5 = 0.333, giving theta2 ≈ 19.1 degrees.
Correct Answer: A — 18.4 degrees
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Q. A light ray passes from diamond (n=2.42) to air. What is the critical angle?
A.
24.4°
B.
30.0°
C.
36.9°
D.
42.0°
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Solution
Critical angle θc = sin⁻¹(1.00/2.42) ≈ 24.4°.
Correct Answer: A — 24.4°
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Q. A light ray passes through the center of curvature of a concave mirror. What will be the angle of reflection?
A.
0 degrees
B.
30 degrees
C.
45 degrees
D.
90 degrees
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Solution
When a light ray passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.
Correct Answer: A — 0 degrees
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Q. A light ray strikes a glass slab at an angle of incidence of 45 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
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Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, sin(r) = (1 * sin(45)) / 1.5 = 0.471, giving r ≈ 28 degrees.
Correct Answer: A — 30 degrees
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Q. A light ray strikes a glass surface at an angle of 60 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?
A.
30 degrees
B.
40 degrees
C.
60 degrees
D.
90 degrees
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Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5 (glass), i = 60 degrees. Solving gives r = 40 degrees.
Correct Answer: B — 40 degrees
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Q. A light ray strikes a glass surface at an angle of incidence of 30 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?
A.
20 degrees
B.
30 degrees
C.
40 degrees
D.
50 degrees
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Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), θ1 = 30 degrees, n2 = 1.5. Thus, sin(θ2) = (1 * sin(30))/1.5 = 1/3, giving θ2 ≈ 20 degrees.
Correct Answer: A — 20 degrees
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Q. A light ray strikes a plane mirror at an angle of 45 degrees. What is the angle of reflection?
A.
0 degrees
B.
45 degrees
C.
90 degrees
D.
30 degrees
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Solution
According to the law of reflection, the angle of reflection equals the angle of incidence, so it is 45 degrees.
Correct Answer: B — 45 degrees
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Q. A light ray traveling in a medium with a refractive index of 1.6 strikes a boundary with air at an angle of 50°. What will be the outcome?
A.
Total internal reflection occurs.
B.
Light is refracted into the air.
C.
Light is absorbed.
D.
Light is scattered.
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Solution
The critical angle for this scenario is approximately 38.7°. Since 50° is greater than the critical angle, total internal reflection occurs.
Correct Answer: A — Total internal reflection occurs.
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Q. A light ray traveling in a medium with n=1.5 strikes the boundary with n=1.0 at 50°. What occurs?
A.
Total internal reflection
B.
Partial reflection and refraction
C.
Complete refraction
D.
None of the above
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Solution
Since 50° is greater than the critical angle (θc ≈ 41.8°), total internal reflection occurs.
Correct Answer: A — Total internal reflection
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Q. A light ray traveling in a medium with n=1.6 strikes the boundary with air at an angle of 50°. What will happen?
A.
Total internal reflection
B.
Partial reflection and refraction
C.
Complete absorption
D.
No reflection
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Solution
Calculate critical angle: θc = sin^(-1)(1/n) = sin^(-1)(1/1.6) ≈ 38.7°. Since 50° > 38.7°, total internal reflection occurs.
Correct Answer: A — Total internal reflection
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Q. A light ray traveling in a medium with n=2.0 strikes a boundary with air at an angle of incidence of 45°. What will be the angle of refraction in air?
A.
22.5°
B.
45°
C.
60°
D.
90°
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Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 2.0, θ1 = 45°, and n2 = 1.0 (air). Thus, 2.0 * sin(45°) = 1.0 * sin(θ2) leads to sin(θ2) = √2, which gives θ2 = 90°.
Correct Answer: D — 90°
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Q. A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
A.
24.4°
B.
36.9°
C.
42.5°
D.
49.5°
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Solution
Using sin(θc) = n2/n1, where n1 = 2.42 (diamond) and n2 = 1.0 (air), we find sin(θc) = 1.0/2.42, leading to θc ≈ 24.4°.
Correct Answer: A — 24.4°
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Q. A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle for total internal reflection?
A.
24.4°
B.
30.0°
C.
36.0°
D.
42.0°
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Solution
Using Snell's law, sin(θc) = n2/n1 = 1.00/2.42. Thus, θc ≈ 24.4°.
Correct Answer: A — 24.4°
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