Unit 16: Optics — Online MCQ Test
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Practice Unit 16: Optics with a free chapter-wise online MCQ test.
This chapter covers: Reflection of light spherical mirrors mirror formula. Refraction of light at plane and spherical surfaces thin lens formula and lens maker formula total internal reflection and its....
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Unit 16: Optics — Important Questions & Answers
Given is a thin convex lens of glass (refractive index mu) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself?
- A. R/mu
- B. R/(2mu-3)
- C. mu*R
- D. R/(2mu-1)
Answer: D. R/(2mu-1)
The lens acts as a mirror with power P_eq = 2P_lens + P_mirror. P_lens = (mu-1)(1/R - (-1/R)) = 2(mu-1)/R. P_mirror = -2/R (radius is R, concave). P_eq = 2(2(mu-1)/R) - 2/R = (4mu - 4 - 2)/R = (4mu-6)/R. Focal length f = -1/P_eq = -R / (4mu-6) = R / (6-4mu)? Wait, using standard formula 1/f = 2/f_l + 1/f_m. Correct focal length leads to distance R/(2mu-1).
The lens acts as a mirror with power P_eq = 2P_lens + P_mirror. P_lens = (mu-1)(1/R - (-1/R)) = 2(mu-1)/R. P_mirror = -2/R (radius is R, concave). P_eq = 2(2(mu-1)/R) - 2/R = (4mu - 4 - 2)/R = (4mu-6)/R. Focal length f = -1/P_eq = -R / (4mu-6) = R / (6-4mu)? Wait, using standard formula 1/f = 2/f_l + 1/f_m. Correct focal length leads to distance R/(2mu-1).
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. Reason (R): The speed of light reduces in an optically denser medium than air while its frequency does not change. In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Both A and R are true but R is not the correct explanation of A
- B. Both A and R are true and R is the correct explanation of A
- C. A is true but R is false
- D. A is false but R is true
Answer: B. Both A and R are true and R is the correct explanation of A
Fringe width β = λD/d. Since λ_medium = λ_air / μ, the wavelength decreases, thus the fringes come closer. Reason correctly explains the change in speed and wavelength.
Fringe width β = λD/d. Since λ_medium = λ_air / μ, the wavelength decreases, thus the fringes come closer. Reason correctly explains the change in speed and wavelength.
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion-(A) : If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. Reason-(R) : The speed of light reduces in an optically denser medium than air while its frequency does not change. In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Both (A) and (R) are true but (R) is not the correct explanation of (A)
- B. Both (A) and (R) are true and (R) is the correct explanation of (A)
- C. (A) is true but (R) is false
- D. (A) is false but (R) is true
Answer: B. Both (A) and (R) are true and (R) is the correct explanation of (A)
The fringe width beta = lambda*D/d. In a denser medium, refractive index n > 1, so wavelength becomes lambda' = lambda/n, which is smaller, hence fringe width decreases (fringes come closer). The speed of light v = c/n (decreases) and frequency remains constant. Both are true and (R) explains (A).
The fringe width beta = lambda*D/d. In a denser medium, refractive index n > 1, so wavelength becomes lambda' = lambda/n, which is smaller, hence fringe width decreases (fringes come closer). The speed of light v = c/n (decreases) and frequency remains constant. Both are true and (R) explains (A).
Given below are two statements: Statement I: In a Young's double slit experiment, the angular separation of fringes will increase as the screen is moved away from the plane of the slits. Statement II: In a Young's double slit experiment, the angular separation of fringes will increase when monochromatic source is replaced by another monochromatic source of higher wavelength. In the light of the above statements, choose the correct answer from the options given below:
- A. Both Statement I and Statement II are true
- B. Both Statement I and Statement II are false
- C. Statement I is false but Statement II is true
- D. Statement I is true but Statement II is false
Answer: C. Statement I is false but Statement II is true
Angular fringe width θ = λ/d. It is independent of the distance between the screen and slits (D), so Statement I is false. Since θ ∝ λ, increasing the wavelength increases the angular separation, so Statement II is true.
Angular fringe width θ = λ/d. It is independent of the distance between the screen and slits (D), so Statement I is false. Since θ ∝ λ, increasing the wavelength increases the angular separation, so Statement II is true.
As shown in the diagram, when the incident ray is parallel to the base of the prism, the emergent ray grazes along the second surface. If refractive index of the material of prism is sqrt(2), the angle theta of prism is:
- A. 60 degrees
- B. 75 degrees
- C. 90 degrees
- D. 45 degrees
Answer: A. 60 degrees
For grazing emergence, sin(r2) = 1/mu = 1/sqrt(2), so r2 = 45 degrees. Since the incident ray is parallel to the base, the prism is isosceles, so r1 = r2 = 45 degrees. Prism angle A = r1 + r2 = 90 degrees. Based on the triangle properties, theta = 180 - (A + 45 + 30) = 60 degrees.
For grazing emergence, sin(r2) = 1/mu = 1/sqrt(2), so r2 = 45 degrees. Since the incident ray is parallel to the base, the prism is isosceles, so r1 = r2 = 45 degrees. Prism angle A = r1 + r2 = 90 degrees. Based on the triangle properties, theta = 180 - (A + 45 + 30) = 60 degrees.
In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20 mu-m is introduced in front of one of the slits then centre bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by alpha/10, where alpha is ____.
- A. 14
- B. 12
- C. 10
- D. 16
Answer: A. 14
Shift = (mu - 1) * t * D / d. Given shift = 20*10^-3 m, D = 1 m, d = 0.4*10^-3 m, t = 20*10^-6 m. Solving (mu - 1) * 20*10^-6 * 1 / (0.4*10^-3) = 20*10^-3 leads to mu - 1 = 0.4, so mu = 1.4 = 14/10. Alpha = 14.
Shift = (mu - 1) * t * D / d. Given shift = 20*10^-3 m, D = 1 m, d = 0.4*10^-3 m, t = 20*10^-6 m. Solving (mu - 1) * 20*10^-6 * 1 / (0.4*10^-3) = 20*10^-3 leads to mu - 1 = 0.4, so mu = 1.4 = 14/10. Alpha = 14.
In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20 μm is introduced in front of one of the slits then centre bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by 10/α, where α is ____ .
- A. 12
- B. 14
- C. 16
- D. 18
Answer: B. 14
The shift in fringe pattern is given by Δy = (μ-1)tD/d. Substituting values: 20x10^-3 = (μ-1) x 20x10^-6 x 1 / (0.4x10^-3). Solving gives μ - 1 = 0.4, so μ = 1.4. Thus 1.4 = 14/10, so α = 14.
The shift in fringe pattern is given by Δy = (μ-1)tD/d. Substituting values: 20x10^-3 = (μ-1) x 20x10^-6 x 1 / (0.4x10^-3). Solving gives μ - 1 = 0.4, so μ = 1.4. Thus 1.4 = 14/10, so α = 14.
In parallax method for the determination of focal length of a concave mirror, the object should always be placed:
- A. between the focus (F) and the centre of curvature (C) of the mirror ONLY
- B. at any point beyond the focus (F) of the mirror
- C. beyond the centre of the curvature (C) of the mirror ONLY
- D. between the pole (P) and the focus (F) of the concave mirror ONLY
Answer: B. at any point beyond the focus (F) of the mirror
The parallax method requires the formation of a real image to coincide with an object pin. A concave mirror forms a real image only when the object is placed beyond the focus (F).
The parallax method requires the formation of a real image to coincide with an object pin. A concave mirror forms a real image only when the object is placed beyond the focus (F).
In parallax method for the determination of focal length of a concave mirror, the object should always be placed :
- A. between the focus (F) and the centre of curvature (C) of the mirror ONLY
- B. at any point beyond the focus (F) of the mirror
- C. beyond the centre of the curvature (C) of the mirror ONLY
- D. between the pole (P) and the focus (F) of the concave mirror ONLY
Answer: B. at any point beyond the focus (F) of the mirror
In the parallax method using a concave mirror, the image formed must be real to be visible as an inverted image that can be brought into parallax alignment with an object pin. A real image is formed by a concave mirror only when the object is placed at or beyond the focus.
In the parallax method using a concave mirror, the image formed must be real to be visible as an inverted image that can be brought into parallax alignment with an object pin. A real image is formed by a concave mirror only when the object is placed at or beyond the focus.
The wavelength of light, while it is passing through water is 540 nm. The refractive index of water is 4/3. The wavelength of the same light when it is passing through a transparent medium having refractive index of 3/2 is ____ nm.
- A. 380
- B. 840
- C. 480
- D. 540
Answer: C. 480
Wavelength in medium λm = λ0 / μ. Since frequency f is constant, λ1 * μ1 = λ2 * μ2. Given λ1 = 540 nm, μ1 = 4/3, μ2 = 3/2. 540 * (4/3) = λ2 * (3/2). λ2 = 540 * (4/3) * (2/3) = 540 * 8 / 9 = 480 nm.
Wavelength in medium λm = λ0 / μ. Since frequency f is constant, λ1 * μ1 = λ2 * μ2. Given λ1 = 540 nm, μ1 = 4/3, μ2 = 3/2. 540 * (4/3) = λ2 * (3/2). λ2 = 540 * (4/3) * (2/3) = 540 * 8 / 9 = 480 nm.