JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 27)
The graph shows how the magnification m produced by a thin lens varies with image distance v. What is the focal length of the lens used?
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$${{{b^2}} \over {ac}}$$
$${{{b^2}c} \over a}$$
$${a \over c}$$
$${b \over c}$$
Explanation
As the graph between magnification (m) and
image distance (v) varies linearly, then
m = k1v + k2
$$ \Rightarrow {v \over u} = {k_1}v + {k_2}$$
$$ \Rightarrow {1 \over u} = {k_1} + {{{k_2}} \over v}$$
$$ \Rightarrow {{{k_2}} \over v} - {1 \over u} = {k_1}$$
Clearly, $${k_1} = {1 \over f}$$ and k2 = 1 here
$$ \therefore f = {1 \over {slope\,of\,m - y\,graph}} = {b \over c}$$
m = k1v + k2
$$ \Rightarrow {v \over u} = {k_1}v + {k_2}$$
$$ \Rightarrow {1 \over u} = {k_1} + {{{k_2}} \over v}$$
$$ \Rightarrow {{{k_2}} \over v} - {1 \over u} = {k_1}$$
Clearly, $${k_1} = {1 \over f}$$ and k2 = 1 here
$$ \therefore f = {1 \over {slope\,of\,m - y\,graph}} = {b \over c}$$
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