Question 189: Ray Optics

Question

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.

Accepted Answer

Given, initially refractive index (n
1
) = 1.4
Then it placed in medium of same refractive index.
$$ 	herefore $$ n
2
= 1.4
We know, Focal length
$${1 \over f} = \left( {{{{n_1}} \over {{n_2}}} - 1} ight)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} ight)$$
$$ \Rightarrow $$ $${1 \over f} = \left( {{{1.4} \over {1.4}} - 1} ight)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} ight)$$
$$ \Rightarrow $$ $${1 \over f} = 0$$
$$ \Rightarrow $$ f = infinite

Answer

Zero

Answer

Infinite

Answer

1

Answer

$${{{R_1}{R_2}} \over {{R_1} - {R_2}}}$$

Ray Optics

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.

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