Question

Find the optical power and the focal lengths

(a)

of a thin glass lens, in a liquid with refractive index

n_{0} = 1.7

and refractive index of glass in air is

1.5

, if its optical power in air is

Φ_{0} = - 5.0 D

;

(b)

of a thin symmetrical biconvex glass lens, with air on one side and water on the other side, if the optical power of that lens in air is

Φ_{0} = + 10 D

.

Accepted Answer

$(a)$ Optical power of a thin lens of refractive index $n_{0} = 1.7$ , in a medium with refractive index $n_{0}$ , is given by,

$Φ = \frac{1}{f} = (n - n_{0}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ ... $(A)$

From equation $(A),$ when the lens is placed in air,

$Φ_{0} = (n - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ ... $(1)$

Similarly, from equation $(A)$ , when the lens is placed in liquid,

$Φ = (n - n_{0}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ ... $(2)$

Thus, from equations $(1)$ and $(2)$ ,

$Φ = (\frac{n - n_{0}}{n - 1}) Φ_{0} = (\frac{1.5 - 1.7}{1.5 - 1}) \times (- 5) = 2 D ∵_{a} n_{g} = 1.5$

Second focal length is given by,

$f' = \frac{n'}{Φ}$ , where $n'$ is the refractive index of the medium in which it is placed.

$f' = \frac{n_{0}}{Φ} = \frac{1.7}{2} = 0.85 m = 85 cm$

$(b)$ Optical power of a thin lens of refractive index $n$ , placed in a medium of refractive index $n_{0}$ , is given by,

$Φ = (n - n_{0}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ ... $(A)$

For a biconvex lens placed in air, from equation $(A)$ ,

$Φ_{0} = (n - 1) (\frac{1}{R} - \frac{1}{- R}) = \frac{2 (n - 1)}{R}$ ... $(1)$

where $R$ is the radius of each curved surface of the lens.

Optical power of a spherical refractive surface is given by,

$Φ = \frac{n' - n}{R}$ ... $(B)$

For the rear surface of the lens which divides air and glass medium,

$Φ_{0} = \frac{n - 1}{R}$ (Here $n$ is the refractive index $(2)$ of glass)

Similarly, for the front surface which divides water and glass medium,

$Φ_{l} = \frac{n_{0} - n}{- R} = \frac{n - n_{0}}{R}$ ... $(3)$

Hence, optical power of the given optical system,

$Φ = Φ_{a} + Φ_{l} = \frac{n - 1}{R} + \frac{n - n_{0}}{R} = \frac{2 n - n_{0} - 1}{R}$ ... $(4)$

From equations $(1)$ and $(4)$ ,

$\frac{Φ}{Φ_{0}} = \frac{2 n - n_{0} - 1}{2 (n - 1)}$ .

So, $Φ = \frac{(2 n - n_{0} - 1)}{2 (n - 1)} Φ_{0} = \frac{(2 \times 1.5 - 1.7 - 1)}{2 (1.5 - 1)} \times (10) = 3 D$

Focal length in air, $f = \frac{1}{Φ} = \frac{1}{3} m = 33.33 cm$

and focal length in water $= \frac{n_{0}}{Φ} = \frac{4}{3 \times 3} = \frac{4}{9} m = 44.44 cm$ , for $n_{0} = \frac{4}{3}$ .

I E Irodov Solutions for Chapter: OPTICS, Exercise 1: PHOTOMETRY AND GEOMETRICAL OPTICS

I E Irodov Physics Solutions for Exercise - I E Irodov Solutions for Chapter: OPTICS, Exercise 1: PHOTOMETRY AND GEOMETRICAL OPTICS

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