Optics
Q. A ray of light 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 ray of light 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 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 reflects back
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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. A ray of light strikes a plane mirror at an angle of 30 degrees. What is the angle of reflection?
A.
30 degrees
B.
60 degrees
C.
90 degrees
D.
45 degrees
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Solution
According to the law of reflection, the angle of reflection is equal to the angle of incidence, which is 30 degrees.
Correct Answer: A — 30 degrees
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Q. A ray of light strikes a plane mirror at an angle of incidence of 30 degrees. What is the angle of reflection?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
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Solution
According to the law of reflection, the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is 30 degrees.
Correct Answer: A — 30 degrees
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Q. A ray of light traveling in air strikes the surface of water at an angle of incidence of 45 degrees. 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). For air (n1=1) and water (n2=1.33), we find r = 45 degrees.
Correct Answer: B — 45 degrees
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Q. A thin lens has a focal length of 20 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 = 1/0.2 = +5 D.
Correct Answer: B — +5 D
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Q. An object is placed 10 cm in front of a convex mirror with a focal length of 5 cm. What is the nature of the image formed?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
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Solution
Convex mirrors always produce virtual and erect images.
Correct Answer: B — Virtual and erect
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Q. An object is placed 25 cm from a convex lens of focal length 10 cm. Where is the image formed?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
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Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 20 cm.
Correct Answer: C — 20 cm
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Q. An object is placed 30 cm from a concave lens of focal length 15 cm. What is the nature of the image formed?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
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Solution
For a concave lens, the image formed is virtual and erect when the object is placed in front of it.
Correct Answer: B — Virtual and erect
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Q. An object is placed 40 cm from a convex lens with a focal length of 10 cm. Where is the image formed?
A.
At 10 cm
B.
At 20 cm
C.
At 30 cm
D.
At 40 cm
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Solution
Using the lens formula, the image is formed at 30 cm on the opposite side.
Correct Answer: C — At 30 cm
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Q. An object is placed 50 cm from a converging lens of focal length 25 cm. Where will the image be formed?
A.
16.67 cm
B.
33.33 cm
C.
25 cm
D.
20 cm
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Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 33.33 cm.
Correct Answer: B — 33.33 cm
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Q. An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Where is the image formed?
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
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Solution
Using the lens formula, 1/f = 1/v - 1/u, we find v = 30 cm, meaning the image is formed 30 cm on the opposite side.
Correct Answer: D — 20 cm
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Q. An object is placed at a distance of 30 cm from a convex lens of focal length 15 cm. What is the distance of the image from the lens?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
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Solution
Using the lens formula, 1/f = 1/v - 1/u. Here, f = 15 cm, u = -30 cm. Solving gives v = 10 cm.
Correct Answer: C — 20 cm
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Q. An object is placed at a distance of 40 cm from a convex lens of focal length 20 cm. Where will the image be formed?
A.
10 cm
B.
20 cm
C.
30 cm
D.
40 cm
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Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 10 cm.
Correct Answer: A — 10 cm
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Q. An object is placed at the focus of a concave lens. 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
A concave lens always forms a virtual and erect image regardless of the object position.
Correct Answer: B — Virtual and erect
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Q. An object is placed at the focus of a concave mirror. What type of image is formed?
A.
Real and inverted
B.
Virtual and upright
C.
No image
D.
Real and upright
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Solution
When the object is at the focus of a concave mirror, the rays are parallel after reflection, resulting in no image being formed.
Correct Answer: C — No image
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Q. For a diffraction grating with 500 lines per mm, what is the angle of the first order maximum for light of wavelength 600 nm?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
15 degrees
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Solution
Using the grating equation d sin θ = nλ, where d = 1/500000 m, n = 1, and λ = 600 x 10^-9 m, we find θ ≈ 30 degrees.
Correct Answer: A — 30 degrees
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Q. For a diffraction pattern produced by a single slit, how does the width of the central maximum compare to the other maxima?
A.
Wider than all other maxima
B.
Narrower than all other maxima
C.
Equal to all other maxima
D.
None of the above
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Solution
The central maximum in a single-slit diffraction pattern is wider than all other maxima.
Correct Answer: A — Wider than all other maxima
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Q. For a diffraction pattern produced by a single slit, how does the width of the central maximum change if the slit width is halved?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
If the slit width is halved, the width of the central maximum increases because the angle for the first minimum increases.
Correct Answer: A — Increases
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Q. For a single slit of width 'a', what is the angular position of the first minimum?
A.
λ/a
B.
a/λ
C.
sin θ = λ/a
D.
tan θ = λ/a
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Solution
The angular position of the first minimum is given by sin θ = λ/a.
Correct Answer: C — sin θ = λ/a
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Q. For destructive interference to occur in a thin film, the path difference must be equal to:
A.
nλ/2 (n is an integer)
B.
nλ (n is an integer)
C.
λ/4
D.
λ/2
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Solution
Destructive interference occurs when the path difference is an odd multiple of λ/2 (i.e., (2n+1)λ/2).
Correct Answer: A — nλ/2 (n is an integer)
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Q. If a beam of light passes through a prism with an angle of 60 degrees and the refractive index of the prism is √3, what is the angle of minimum deviation?
A.
30 degrees
B.
60 degrees
C.
45 degrees
D.
15 degrees
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Solution
Using the formula for minimum deviation, D = (n - 1)A. Here, n = √3 and A = 60 degrees. Thus, D = (√3 - 1) * 60 degrees, which approximates to 30 degrees.
Correct Answer: A — 30 degrees
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Q. If a concave lens has a focal length of -10 cm, what is the nature of the image formed when an object is placed at 5 cm from the lens?
A.
Real and inverted
B.
Virtual and upright
C.
Real and upright
D.
Virtual and inverted
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Solution
Concave lenses always produce virtual and upright images.
Correct Answer: B — Virtual and upright
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Q. If a concave lens has a focal length of -10 cm, what is the nature of the image formed when the object is placed at 5 cm?
A.
Real and inverted
B.
Virtual and upright
C.
Real and upright
D.
Virtual and inverted
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Solution
Concave lenses always produce virtual and upright images.
Correct Answer: B — Virtual and upright
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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 lens has a focal length of -20 cm, what is the nature of the image formed when an object is placed at 30 cm from the lens?
A.
Real and inverted
B.
Virtual and upright
C.
Real and upright
D.
Virtual and inverted
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Solution
For a concave lens, the image formed is virtual and upright when the object is placed beyond the focal length.
Correct Answer: B — Virtual and upright
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Q. If a concave lens has a focal length of -20 cm, what is the nature of the image formed when an object is placed at 30 cm?
A.
Real and inverted
B.
Virtual and upright
C.
Real and upright
D.
Virtual and inverted
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Solution
The image formed by a concave lens is virtual and upright when the object is placed beyond the focal length.
Correct Answer: B — Virtual and upright
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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 forms a real image at a distance of 40 cm from the lens, and the object is placed at 60 cm, what is the focal length of the lens?
A.
15 cm
B.
20 cm
C.
25 cm
D.
30 cm
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Solution
Using the lens formula 1/f = 1/v - 1/u, we have 1/f = 1/40 - 1/(-60) = 1/40 + 1/60 = 1/24. Therefore, f = 24 cm.
Correct Answer: C — 25 cm
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