Optics 3 Question 39
39. A plano-convex lens has a thickness of $4 cm$. When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be $3 cm$. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face is found to be $25 / 8 cm$. Find the focal length of the lens. Assume thickness to be negligible while finding its focal length.
(1984, 6M)
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Answer:
Correct Answer: 39. $\mu _1<\mu _2$
Solution:
- Refer figure (a)
(a)
(b)
In this case refraction of the rays starting from $O$ takes place from a plane surface. So, we can use
$$ d _{\text {app }}=\frac{d _{\text {actual }}}{\mu} \quad \text { or } \quad 3=\frac{4}{\mu} \quad \text { or } \quad \mu=\frac{4}{3} $$
Refer figure (b) In this case refraction takes place from a spherical surface. Hence, applying
$$ \begin{array}{rlrl} & \frac{\mu _2}{v}-\frac{\mu _1}{u}=\frac{\mu _2-\mu _1}{R} \\ \text { we have, } & \frac{1}{(-25 / 8)}-\frac{4 / 3}{-4}=\frac{1-4 / 3}{-R} \\ \text { or } & \frac{1}{3 R} & =\frac{1}{3}-\frac{8}{25}=\frac{1}{75} \\ \therefore & R & =25 cm \end{array} $$
Now, to find the focal length we will use the lens Maker’s formula
$$ \begin{aligned} \frac{1}{f} & =(\mu-1) \frac{1}{R _1}-\frac{1}{R _2}=\frac{4}{3}-1 \frac{1}{\infty}-\frac{1}{-25}=\frac{1}{75} \\ \therefore \quad f & =75 cm \end{aligned} $$
