## Question

The thickness of a plano-convex lens is 4 cm. When it is placed on a horizontal table in such a way that its curved surface be in contact with the table, then the depth of a bottom-point of the lens appears 3 cm. If the lens is inverted so that its plane surface be in contact with the table, then the apparent depth of the centre of the lens-surface is found to be (25/8) cm. Determine the focal-length of the lens.

### Solution

75 cm

In the first position, the curved surface of the lens is in contact with the table. Let the contact-point *O* be an object. The light rays starting from *O*, after refraction in air at the plane surface of the lens, forms virtual image *I*of the object *O*. Now, as given, we have

Putting these values in the formula for refraction at a surface

Here the refracting surface is plane whose radius *R* = ∞.

or *n* = 3/4.

Since light-rays are going form glass to air; here *n* is refractive index of air with respect to glass (*n* = * _{g}n_{a}* = 3/4).

In the second position, the plane surface of the lens is in contact with the table. Now, the virtual image *I* of the centre *O* of the plane surface of the lens is formed due to refraction at the curved surface of the lens. Again, as given

*u* = *CO* = – 4 cm and *v* = *CI* = –(25/8) cm and *n* = 3/4

(calculated above)

Let *R* be the radius of the curved surface of the lens.

Putting the values of *u*, *v* and *n* in the formula

or

or

or *R* = –25 cm.

The focal length *f* of the lens in air is given by

Where *n* is refractive index of glass of lens with respect to air.

Here

or *f* = 75 cm.

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