The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R1 and R2 respectively.
Solution
Given, initially refractive index (n<sub>1</sub>) = 1.4
<br><br>Then it placed in medium of same refractive index.
<br><br>$\therefore$ n<sub>2</sub> = 1.4
<br><br>We know, Focal length
<br><br>$${1 \over f} = \left( {{{{n_1}} \over {{n_2}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
<br><br>$\Rightarrow$ $${1 \over f} = \left( {{{1.4} \over {1.4}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
<br><br>$\Rightarrow$ ${1 \over f} = 0$
<br><br>$\Rightarrow$ f = infinite
About this question
Subject: Physics · Chapter: Optics · Topic: Lenses and Optical Instruments
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