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What is the focal length of a double convex lens kept in air with two spherical surfaces of radii ${R_1} = 30cm$ and ${R_2} = 60cm$? (Take the refractive index of lens as 1.5)

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Answer
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- Hint- We will use the Lens Maker Formula in order to find the focal length as per asked by the question. The formula is $\dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$. Where, n is the refractive index of the lens given in the question.

Complete step-by-step solution -

We will make a figure for better understanding-

The formula used to find focal length of a convex lens is as follows-
$ \Rightarrow \dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where n stands for refractive index which as per the question is 1.5.
The value of ${R_1}$ as per given in the question is 30cm and the value of ${R_2}$ as per given in the question is 60cm.
Putting all these values in the formula we get-
$ \Rightarrow \dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
$ \Rightarrow \dfrac{1}{f} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{30}} - \dfrac{1}{{\left( { - 60} \right)}}} \right)$ (since $R_2$ is negative)
$
   \Rightarrow \dfrac{1}{f} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{30}} - \dfrac{1}{{\left( { - 60} \right)}}} \right) \\
    \\
   \Rightarrow \dfrac{1}{f} = 0.5\left( {\dfrac{1}{{30}} + \dfrac{1}{{60}}} \right) \\
    \\
   \Rightarrow \dfrac{1}{f} = 0.5\left( {\dfrac{{60 + 30}}{{60 \times 30}}} \right) \\
    \\
   \Rightarrow f = \dfrac{{60 \times 30}}{{0.5 \times \left( {60 + 30} \right)}} \\
    \\
   \Rightarrow f = \dfrac{{1800}}{{0.5 \times 90}} \\
    \\
   \Rightarrow f = 40cm \\
$
Thus, the focal length of the given double convex lens as per asked by the question is 40cm.

Note: Remember that the value is negative for in the question above and do not forget to put the negative sign or the answer may vary. Such questions can be solved by using the simple Lens Maker Formula.