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Glass-slab refractive index using travelling microscope: NEET MCQs with answers

50 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In a travelling microscope experiment, three readings on the vertical scale are taken: R₁ when focused directly on a mark on a sheet of paper, R₂ when focused on the mark through a transparent slab placed over it, and R₃ when focused on chalk dust sprinkled on the top surface of the slab. If R₁ < R₂ < R₃, the refractive index (μ) of the slab is given by:

  1. (R₂ - R₁) / (R₃ - R₂)
  2. (R₃ - R₂) / (R₃ - R₁)
  3. (R₃ - R₁) / (R₃ - R₂)
  4. (R₃ - R₁) / (R₂ - R₁)
Show answer and explanation

Answer: (C) (R₃ - R₁) / (R₃ - R₂)

The bottom surface of the slab corresponds to the mark on paper (reading R₁), the apparent image of the mark seen through the slab is at reading R₂, and the top surface of the slab is at reading R₃. Therefore, the real thickness of the slab is (R₃ - R₁) and the apparent thickness is (R₃ - R₂). The refractive index μ is real thickness divided by apparent thickness, giving μ = (R₃ - R₁) / (R₃ - R₂).

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2. In a travelling microscope experiment, why are only paraxial rays allowed to enter the objective of the microscope?

  1. To minimize spherical aberration so that a sharp image is formed at a single depth
  2. To eliminate chromatic aberration caused by using white light
  3. To prevent total internal reflection at the upper surface of the glass slab
  4. To maximize the brightness and magnification of the ink dot
Show answer and explanation

Answer: (A) To minimize spherical aberration so that a sharp image is formed at a single depth

Refraction at a plane surface gives an apparent depth d' = t/μ strictly for paraxial rays (small angles of incidence). Marginal rays undergo greater refraction causing spherical aberration, which blurs the image and shifts its apparent position.

hard

3. A glass slab of refractive index 1.5 is placed over a paper with a blue dot. If the experiment is repeated by replacing the blue dot with a red dot (viewed through the same slab and using the same illumination), the apparent shift of the dot will:

  1. Increase
  2. Remain exactly the same
  3. Decrease
  4. Become zero
Show answer and explanation

Answer: (C) Decrease

According to Cauchy's dispersion formula, refractive index decreases as wavelength increases (μ_red < μ_blue). Apparent shift is given by s = t(1 - 1/μ). Since μ_red is smaller than μ_blue, the term (1 - 1/μ) is smaller for red light, which means the apparent shift decreases.

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4. A glass slab is composed of two identical halves fused together along their faces. Each half has a thickness of 2.0 cm. The lower half has a refractive index of 1.60 and the upper half has a refractive index of 1.40. When an ink dot at the bottom is viewed normally through this composite slab, its apparent depth from the top surface is:

  1. 2.68 cm
  2. 2.50 cm
  3. 2.82 cm
  4. 3.12 cm
Show answer and explanation

Answer: (A) 2.68 cm

Apparent depth d' = t₁/n₁ + t₂/n₂ = (2.0 / 1.60) + (2.0 / 1.40) = 1.25 cm + 1.4286 cm ≈ 2.68 cm.

easy

5. In an experiment using a travelling microscope, fine lycopodium powder or chalk dust is sprinkled on the upper surface of the glass slab primarily to:

  1. Increase the friction between the microscope objective and the slab
  2. Provide a distinct opaque object to focus sharply on the top surface of the slab
  3. Prevent the total internal reflection of light inside the transparent slab
  4. Absorb stray light and increase the refractive index of glass
Show answer and explanation

Answer: (B) Provide a distinct opaque object to focus sharply on the top surface of the slab

Because a clean glass surface is completely transparent, the microscope cannot easily distinguish the exact focal plane of the top surface. Scattering particles like lycopodium powder or chalk dust provide sharp, opaque reference marks on that plane.

hard

6. A student measures the refractive index of a transparent medium with a travelling microscope using monochromatic violet light and then repeats the measurement using monochromatic red light. If n_v and n_r are the measured refractive indices and s_v and s_r are the observed upward shifts of the base mark, which relation is correct?

  1. n_v < n_r and s_v > s_r
  2. n_v > n_r and s_v < s_r
  3. n_v > n_r and s_v > s_r
  4. n_v < n_r and s_v < s_r
Show answer and explanation

Answer: (C) n_v > n_r and s_v > s_r

Due to dispersion (Cauchy's relation), optical media have higher refractive indices for shorter wavelengths, so n_v > n_r. The apparent shift is s = t(1 - 1/n). Since n_v > n_r, 1/n_v < 1/n_r, which implies (1 - 1/n_v) > (1 - 1/n_r), meaning s_v > s_r.

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7. A glass slab contains a tiny air bubble. When viewed perpendicularly from one face using a travelling microscope, the bubble appears to be at a depth of 3.0 cm. When viewed perpendicularly from the opposite face, it appears to be at a depth of 5.0 cm. If the actual thickness of the slab is 12.0 cm, the refractive index of the glass is:

  1. 1.45
  2. 1.33
  3. 1.60
  4. 1.50
Show answer and explanation

Answer: (D) 1.50

Let the true distance of the bubble from the two faces be x and 12 - x. Apparent depths are d₁' = x/μ and d₂' = (12 - x)/μ. Adding both gives d₁' + d₂' = 12/μ. Thus 3.0 + 5.0 = 12/μ, which yields μ = 12 / 8.0 = 1.50.

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8. An ink mark is placed on a flat horizontal table. A glass cube of side 6 cm and refractive index 1.5 is placed over the mark. A microscope is focused on the image of the mark. If a parallel-faced transparent liquid layer of thickness 2 cm and refractive index 1.6 is poured on the top face of the glass cube, how far must the microscope be further moved to bring the ink mark into focus again?

  1. 1.25 cm upwards
  2. 0.75 cm downwards
  3. 0.75 cm upwards
  4. 0.40 cm upwards
Show answer and explanation

Answer: (C) 0.75 cm upwards

When an additional layer of liquid of thickness t_L and refractive index μ_L is added, the apparent position of the mark shifts further upwards by Δs = t_L(1 - 1/μ_L) along the line of sight. Here t_L = 2 cm and μ_L = 1.6 = 8/5. The additional shift is Δs = 2(1 - 5/8) = 2(3/8) = 6/8 = 0.75 cm. Thus, the microscope must be moved 0.75 cm upwards.

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9. Microscope readings are R₁ (mark without slab), R₂ (mark seen through slab) and R₃ (top surface of slab). The refractive index is:

  1. (R₃ - R₁)/(R₃ - R₂)
  2. (R₂ - R₁)/(R₃ - R₂)
  3. (R₃ - R₂)/(R₃ - R₁)
  4. R₃/R₁
Show answer and explanation

Answer: (A) (R₃ - R₁)/(R₃ - R₂)

Real thickness = R₃ - R₁ and apparent depth = R₃ - R₂, so n = (R₃ - R₁)/(R₃ - R₂).

NCERT Class 11

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10. A student uses a travelling microscope to measure the refractive index of a rectangular block of thickness 6.0 cm. However, a thin film of transparent oil (refractive index 1.20) of thickness 1.2 cm is poured over the glass slab. If the glass has refractive index 1.50, what is the total vertical shift of an ink dot placed beneath the glass block?

  1. 2.0 cm
  2. 1.8 cm
  3. 2.5 cm
  4. 2.2 cm
Show answer and explanation

Answer: (D) 2.2 cm

Total apparent shift = s₁ + s₂ = t₁(1 - 1/n₁) + t₂(1 - 1/n₂). For the glass slab: s₁ = 6.0 × (1 - 1/1.50) = 6.0 × (1/3) = 2.0 cm. For the oil layer: s₂ = 1.2 × (1 - 1/1.20) = 1.2 × (0.2/1.2) = 0.2 cm. Total shift = 2.0 + 0.2 = 2.2 cm.

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11. In an experiment to determine the refractive index of a glass block, the main scale reading for the apparent position of an ink mark is 5.25 cm with a vernier coincidence of 18 divisions. For the top surface dusted with powder, the main scale reads 6.75 cm with a vernier coincidence of 38 divisions. If the least count of the microscope is 0.001 cm, what is the apparent thickness of the block?

  1. 1.520 cm
  2. 1.500 cm
  3. 1.480 cm
  4. 1.538 cm
Show answer and explanation

Answer: (A) 1.520 cm

Apparent mark reading = 5.25 + (18 × 0.001) = 5.268 cm. Top surface reading = 6.75 + (38 × 0.001) = 6.788 cm. Apparent thickness = 6.788 - 5.268 = 1.520 cm.

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12. While focusing on the ink dot through a glass slab of thickness 6.0 cm, the microscope tube had to be raised by 2.0 cm from its initial position focused on the bare dot. What is the refractive index of the glass material?

  1. 1.33
  2. 1.75
  3. 1.67
  4. 1.50
Show answer and explanation

Answer: (D) 1.50

The apparent shift s = t(1 - 1/μ). Given t = 6.0 cm and shift s = 2.0 cm, we have 2.0 = 6.0(1 - 1/μ). This gives 1 - 1/μ = 1/3, which leads to 1/μ = 2/3, and hence μ = 1.50.

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