Question

3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic

Answer 1

The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

$\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}$

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

$The \ refractive \ index, \ n = \dfrac{d_1}{d_2}$

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

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