JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 2)
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.
Zero
Infinite
1
$${{{R_1}{R_2}} \over {{R_1} - {R_2}}}$$
Explanation
Given, initially refractive index (n1) = 1.4
Then it placed in medium of same refractive index.
$$ \therefore $$ n2 = 1.4
We know, Focal length
$${1 \over f} = \left( {{{{n_1}} \over {{n_2}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$$ \Rightarrow $$ $${1 \over f} = \left( {{{1.4} \over {1.4}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$$ \Rightarrow $$ $${1 \over f} = 0$$
$$ \Rightarrow $$ f = infinite
Then it placed in medium of same refractive index.
$$ \therefore $$ n2 = 1.4
We know, Focal length
$${1 \over f} = \left( {{{{n_1}} \over {{n_2}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$$ \Rightarrow $$ $${1 \over f} = \left( {{{1.4} \over {1.4}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$$ \Rightarrow $$ $${1 \over f} = 0$$
$$ \Rightarrow $$ f = infinite
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