Question

\[\left( i \right)\] if \[f = 0.5m\] for a glass lens. What is the power of the lens \[?\]
\[\left( {ii} \right)\] The radii of curvature of the faces of a double convex lens are \[10cm\] and\[15cm\]. Its focal length is\[12cm\] . What is the refractive index of glass \[?\]
\[\left( {iii} \right)\] a convex lens has \[20cm\] focal length in air. What is the focal length in water \[?\] \[\left( {} \right.\]refractive index of air water \[ = \] \[1.33\], refractive index for air glass \[ = \]\[1.5\]\[\left. {} \right)\]

Answer 1

Hint:Here we have to know all the formulas of power, focal length and radius of curvature. And all the units should be in the same unit either S.I or C.G.S. if they are not in the same units then we have to convert them and convert them. Here the refractive index of water is given. Refractive index of glass is also given. We have to apply those in the calculation.

Complete step by step answer:
\[\left( i \right)\] We know that,\[f = 0.5m\], here \[f = \] focal length of that glass.Now, the power \[p = \dfrac{1}{f}\]. Which is in a meter unit.Now putting the values we get, \[p = \dfrac{1}{{0.5}} = 2D\]. Here the unit of power is equal to Dioptre. So, the power of the glass is \[2D\] .

\[\left( {ii} \right)\] using lens maker formula we can say
\[\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Here \[\mu \]is the refractive index of the convex lens. We have to calculate that.
Now, \[{R_1}\]\[ = \]\[10cm\] and \[{R_2}\]\[ = \] \[ - \]\[15cm\], according to sign convention law. Now putting the value we get,
\[\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\\
\Rightarrow\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{10}} + \dfrac{1}{{15}}} \right) \\
\Rightarrow\dfrac{1}{f} = \left( {\mu - 1} \right) \times \dfrac{1}{6}\\
\Rightarrow \left( {\mu - 1} \right) = 0.5 \\
\Rightarrow \mu = 0.5 + 1 = 1.5 \\ \]
Therefore the refractive index of the convex lens is \[1.5\] .

\[\left( {iii} \right)\] we know from the previous solution the lanes maker formula is,
\[\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
We are fist calculating this for air.
Now,
 \[\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) = \]\[\dfrac{1}{{f\left( {\mu - 1} \right)}}\]\[ = \dfrac{1}{{20\left( {1.5 - 1} \right)}} = \dfrac{1}{{10}}\]
 In case of water
\[\dfrac{1}{{{f^1}}} = \left( {\dfrac{{1.5}}{{1.33}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\
\Rightarrow\dfrac{1}{{{f^1}}} = \left( {\dfrac{{1.55 - 1.33}}{{1.33}}} \right)\left( {\dfrac{1}{{10}}} \right)\]
\[\therefore{f^1} = 78.23cm\]
Focal length in water is \[78.23cm\] .

Note:we all can get confused by the signs of \[{R_1}\] and \[{R_2}\] . We have to carefully select whether it will be positive or negative. And here also one should calculate all the values in the same unit. There will be some questions where glass and air refractive index will not be given. So we all should memorise the value of those refractive indices. here the refractive index of glass is given. The refractive index of water is also given.