Question
Question: An air bubble in a glass with refractive index \(1.5\) (near normal incidence) i.e. \(5\,cm\) deep w...
An air bubble in a glass with refractive index 1.5 (near normal incidence) i.e. 5cm deep when viewed from one surface and 3cm deep when viewed from the opposite face. The thickness is:
(A) 16
(B) 8
(C) 10
(D) 12
Solution
Hint
First by using the given refractive index and the apparent depth value, the real depth is determined. Then by using the second apparent depth value, the second real depth can be determined, by adding the two real depth values, the thickness can be determined.
The refractive index of the air bubble is given by,
⇒μ=d2d1
Where, μ is the refractive index of the air bubble, d1 is the real depth and d2 is the apparent depth.
Complete step by step answer
Given that, The refractive index of the air bubble is, μ=1.5.
The first apparent depth value is, d2=5cm.
The second apparent depth value is, d2=3cm.
Now, The refractive index of the air bubble is given by,
⇒μ=d2d1..................(1)
By substituting the refractive index of the air bubble and the first apparent depth in the above equation (1), then
⇒1.5=5d1
By rearranging the terms in the above equation, then
⇒d1=1.5×5
By multiplying the terms in the above equation, then
⇒d1=7.5cm
Now, substituting the refractive index of the air bubble and the second apparent depth in the above equation (1), then
⇒1.5=3d1
By rearranging the terms in the above equation, then
⇒d1=1.5×3
By multiplying the terms in the above equation, then
⇒d1=4.5cm
The thickness of the air bubble is, ⇒7.5cm+4.5cm.
Then the thickness is 12cm.
Hence, the option (D) is the correct answer.
Note
The refractive index is directly proportional to the real depth and the refractive index is inversely proportional to the apparent depth. As the refractive index increases when the real depth increases and the apparent depth decreases. As the refractive index decreases when the real depth decreases and the apparent depth increases.