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Physics

A scientist lowers a metallic ruler vertically into a transparent oil tank. The ruler touches an object placed at the bottom of the tank and gets wet up to the 25 cm mark. If the refractive index of the glycerin is 1.25:

(a) up to which mark will the ruler get wet, if the scientist lowers it up to the image of the object?

(b) how will this length in (a) change if another liquid of μ > 1.25 is used?

Refraction Plane Surfaces

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Answer

(a) Given,

  • Actual depth of the object = 25 cm
  • Refractive index of the glycerin = μ = 1.25
  • The ruler is seen to be touching the object visually, so it is lowered up to the apparent position of the object (i.e., the image due to refraction)

As,

Apparent depth=Real depthμ=251.25\text {Apparent depth} = \dfrac{\text {Real depth}}{\text μ} = \dfrac{25}{1.25} = 20 cm

So, the ruler will get wet up to the 20 cm mark.

(b) As,

Apparent depth=Real depthμ\text {Apparent depth} = \dfrac{\text {Real depth}}{\text μ}

Here, apparent depth depends inversely on refractive index (μ) of the medium so if another liquid of μ > 1.25 is used then apparent depth decreases.

Hence, the wet length of the ruler will be less than 20 cm.

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