Answer
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Hint: Material with a high index of refraction will have a shorter focal length than those with lower refractive indices. Let us take an example of lenses made of the synthetic polymers which is Lucite which is having a lower refractive index than glass which is leading to a slightly longer focal length.
Complete answer:
From the Lens maker formula,
$
\dfrac{1}{f} = (\mu - 1)\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
\Rightarrow \dfrac{1}{{{f_a}}} = (1.5 - 1)\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
\Rightarrow \dfrac{1}{{{f_1}}} = ({\mu _g} - 1)\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
\Rightarrow {1_{ug}} = \dfrac{{{\mu _g}}}{{{\mu _1}}} = \dfrac{{1.5}}{{1.25}} = \dfrac{6}{5} \\
\Rightarrow \dfrac{1}{{{f_1}}} = \left( {\dfrac{6}{5} - 1} \right)\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) = \dfrac{1}{5}\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
\Rightarrow \dfrac{{1/{f_a}}}{{1/{f_1}}} = \dfrac{{0.5}}{{1/5}} \\
\Rightarrow \dfrac{{{f_1}}}{{{f_a}}} = 0.5 \times 5 = 2.5 \\
\Rightarrow {f_1} = 2.5 \times {f_a} \\
$
Refractive index is said to be the velocity of light in a vacuum to its velocity in a specific medium. In other words it is also defined as the ratio of speed of light in a vacuum to the speed of light. Therefore if we talk about focal length then, the focal length of a lens is inversely proportional to refractive index of material of the lens medium. Moreover, the refractive index (m) increases focal length (f) of the lens decreases.
So, the correct answer is “Option B”.
Note:
There are two kinds of refractive index i-e refractive index from rarer medium and refractive index through denser medium and two types of refractive index : absolute refractive index and relative refractive index.
Complete answer:
From the Lens maker formula,
$
\dfrac{1}{f} = (\mu - 1)\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
\Rightarrow \dfrac{1}{{{f_a}}} = (1.5 - 1)\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
\Rightarrow \dfrac{1}{{{f_1}}} = ({\mu _g} - 1)\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
\Rightarrow {1_{ug}} = \dfrac{{{\mu _g}}}{{{\mu _1}}} = \dfrac{{1.5}}{{1.25}} = \dfrac{6}{5} \\
\Rightarrow \dfrac{1}{{{f_1}}} = \left( {\dfrac{6}{5} - 1} \right)\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) = \dfrac{1}{5}\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
\Rightarrow \dfrac{{1/{f_a}}}{{1/{f_1}}} = \dfrac{{0.5}}{{1/5}} \\
\Rightarrow \dfrac{{{f_1}}}{{{f_a}}} = 0.5 \times 5 = 2.5 \\
\Rightarrow {f_1} = 2.5 \times {f_a} \\
$
Refractive index is said to be the velocity of light in a vacuum to its velocity in a specific medium. In other words it is also defined as the ratio of speed of light in a vacuum to the speed of light. Therefore if we talk about focal length then, the focal length of a lens is inversely proportional to refractive index of material of the lens medium. Moreover, the refractive index (m) increases focal length (f) of the lens decreases.
So, the correct answer is “Option B”.
Note:
There are two kinds of refractive index i-e refractive index from rarer medium and refractive index through denser medium and two types of refractive index : absolute refractive index and relative refractive index.
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