A thin convex lens is made of a material of refractive index 1.6. An object is kept at a distance of a from lens on the principal axis as shown in the figure. The radius of curvature of the surface are 10 cm and 5 cm. Now, the lens is reversed such that the face having radius of curvature 5 cm lies close to the object. The difference in image position as obtained for both the cases is equal to:-

Question

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Answer 1

$calculating focus in first case \to R_{1} = 10 c m R_{2} = 5 c m μ = refractive index$
using formula
$\frac{1}{f} \frac{4}{f} \frac{1}{f} f = (\frac{μ}{1} - 4) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (1.6 - 1) (\frac{1}{1 0} - \frac{1}{- 5}) = (0.6) [\frac{1}{1 0} + \frac{1}{5}) = 50 / 9 c m$
calculating focus in second case $\to$
$R_{1} = 5 c m R_{2} = 10 c m$
$\frac{1}{f} = (0.6) [\frac{1}{5} + \frac{1}{1 0}]$
$f = \frac{5 0}{9} c m$
The focus of lens
doen't change by reversing
its orientation.
So the difference in image position=0
Hence option ' $D$ is correct.

Answer 2

calculating focus in first case \to R_{1} = 10 c m R_{2} = 5 c m μ = refractive index

using formula

\frac{1}{f} \frac{4}{f} \frac{1}{f} f = (\frac{μ}{1} - 4) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (1.6 - 1) (\frac{1}{1 0} - \frac{1}{- 5}) = (0.6) [\frac{1}{1 0} + \frac{1}{5}) = 50 / 9 c m

calculating focus in second case

\to

R_{1} = 5 c m R_{2} = 10 c m

\frac{1}{f} = (0.6) [\frac{1}{5} + \frac{1}{1 0}]

f = \frac{5 0}{9} c m

The focus of lens

doen't change by reversing

its orientation.

So the difference in image position=0

Hence option '

D

is correct.