Two similar thin equi-convex lenses, of focal length each, are kept coaxially in contact with each other such that the focal length of the combination is ?. When the space between the two lenses is filled with glycerin (which has the same refractive ind

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Solution

Q.
Two similar thin equi-convex lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_{1}$ .

When the space between the two lenses is filled with glycerin (which has the same refractive index $(μ = 1.5)$ as that of glass) then the equivalent focal length is $F_{2}$ . The ratio $F_{1} : F_{2}$ will be [2019]

1 3 : 4

2 2 : 1

3 1 : 2

4 2 : 3

Ans.
(3)

According to lens maker’s formula

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = (1.5 - 1) (\frac{2}{R}) = \frac{1}{R}$

Two similar equi-convex lenses of focal length $f$ each are held in contact with each other.

The focal length $F_{1}$ of the combination is given by

$\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}; F_{1} = \frac{f}{2} = \frac{R}{2}$ ...(i)

For glycerin in between lenses, there are three lenses, one concave and two convex.

Focal length of the concave lens is given by

$\frac{1}{f'} = (1.5 - 1) (\frac{- 2}{R}) = - \frac{1}{R}$

Now, equivalent focal length of the combination is,

$\frac{1}{F_{2}} = \frac{1}{f} + \frac{1}{f'} + \frac{1}{f}; \frac{1}{F_{2}} = \frac{1}{R} - \frac{1}{R} + \frac{1}{R} = \frac{1}{R}$

$F_{2} = R$ ...(ii)

Dividing equation (i) by (ii), we get $\frac{F_{1}}{F_{2}} = \frac{1}{2}$

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