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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Dispersion and Spectra

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Dispersion and Spectra

As the colour of light varies from red to violet, the refractive...

As the colour of light varies from red to violet, the refractive index of a certain material changes by 0.0140.014. A rectangular slab of height 2.00 cm2.00 \text{ cm} constructed from this material is placed over a newspaper. When observed normally in yellow light, the letters appear to be 1.32 cm1.32 \text{ cm} below the top surface of the slab. Determine the dispersive power of the material.

Options

  1. A0.0180.018
  2. B0.0270.027
  3. C0.0350.035
  4. D0.0260.026

Correct answer

B. 0.0270.027

Step-by-step solution

The refractive index for yellow light is given by μy=Real depthApparent depth\mu_y = \dfrac{\text{Real depth}}{\text{Apparent depth}}.

Substituting the given values:

μy=2.001.32=5033\mu_y = \dfrac{2.00}{1.32} = \dfrac{50}{33}

The dispersive power ω\omega is given by the formula:

ω=μv−μrμy−1\omega = \dfrac{\mu_v - \mu_r}{\mu_y - 1}

Given that the change in refractive index from red to violet is μv−μr=0.014\mu_v - \mu_r = 0.014.

Substituting the values into the formula for dispersive power:

ω=0.0145033−1\omega = \dfrac{0.014}{\dfrac{50}{33} - 1}

ω=0.0141733\omega = \dfrac{0.014}{\dfrac{17}{33}}

ω=0.014×3317=0.46217≈0.027\omega = \dfrac{0.014 \times 33}{17} = \dfrac{0.462}{17} \approx 0.027
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