First slide

Ray optics

Question

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

Difficult

A

3 : 4
B

2 : 1
C

1 : 2
D

2 : 3

Solution

$According to lens maker's formula \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = (1.5 - 1) (\frac{2}{R}) = \frac{1}{R}$
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 curve lens is given
$\frac{1}{f^{'}} = (1.5 - 1) (\frac{- 2}{R}) = - \frac{1}{R}$
Now, equivalent focal length of the combination is given by
$\begin{array}{l} \frac{1}{F_{2}} = \frac{1}{f} + \frac{1}{f^{'}} + \frac{1}{f}; \frac{1}{F_{2}} = \frac{1}{R} - \frac{1}{R} + \frac{1}{R} = \frac{1}{R} \\ F_{2} = R . . . . . . . . . . (ii) \end{array} Dividing equation (i) by (ii), we get \frac{F_{1}}{F_{2}} = \frac{1}{2}$

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