Question

A plano convex lens of focal length $20cm$ has its plane side silvered (This question has multiple correct options)
A) The radius of curvature of the curved surface of a given plano-convex lens is equal to half of the radius of curvature of an equiconvex lens of focal length $20cm$.
B) An object placed at $15cm$ on the axis on the convex side of a silvered plano-convex lens gives rise to an image at a distance of $30cm$ from it.
C) An object placed at a distance of $20cm$ on the axis on the convex side of a silvered plano-convex lens gives rise to an image at $40cm$ from it.
D) Silvered plano-convex lens acts as a concave mirror of focal length $10cm$.

Answer 1

Hint: This problem can be solved by using lensmaker’s formula and plugging in the radius values for a plano convex and equiconvex lens, to get the relation between their radii of curvature. The direct formula for the focal length of a concave mirror made from a silvered plano convex lens can be used to solve the next part and finally the lens formula can be used to find the position of the image.

Formula used:
$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$ [Lensmaker’s formula]
$\dfrac{1}{f}=\dfrac{1}{{{f}_{mirror}}}-\dfrac{2}{{{f}_{lens}}}$
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

Complete step-by-step answer:
Let the radius of curvature of the curved surface of the plano convex lens be ${{R}_{plano}}$ while the radius of curvature of the equiconvex lens be ${{R}_{eq}}$.
Let the focal lengths of the planoconvex and equiconvex lenses be ${{f}_{plano}}$ and ${{f}_{eq}}$ respectively.
Now, the lensmaker’s formula says that the focal length $f$ of a lens made of material with refractive index $\mu $ and kept in air is given by
$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$ --(1)
Where ${{R}_{1}},{{R}_{2}}$ are the radii of curvatures of the first and second curved surfaces respectively.
Therefore, using (1),
We get
$\dfrac{1}{{{f}_{plano}}}=\left( \mu -1 \right)\left( \dfrac{1}{+{{R}_{plano}}}-\dfrac{1}{\infty } \right)=\left( \mu -1 \right)\left( \dfrac{1}{+{{R}_{plano}}}-0 \right)=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{plano}}} \right)$ (Since, for a plane surface, radius of curvature is $\infty $) –(2)
Also, using (1), we get,
$\dfrac{1}{{{f}_{eq}}}=\left( \mu -1 \right)\left( \dfrac{1}{+{{R}_{eq}}}-\dfrac{1}{-{{R}_{eq}}} \right)=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{eq}}}+\dfrac{1}{{{R}_{eq}}} \right)=\left( \mu -1 \right)\left( \dfrac{2}{{{R}_{eq}}} \right)$ -(3) (Since for an equiconvex lens ${{R}_{1}}={{R}_{2}}$ and the negative sign is due to sign convention)
Now, if ${{f}_{plano}}={{f}_{eq}}$ $\therefore \dfrac{1}{{{f}_{plano}}}=\dfrac{1}{{{f}_{eq}}}$
Therefore, using (2) and (3), we will get,
$\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{plano}}} \right)=\left( \mu -1 \right)\left( \dfrac{2}{{{R}_{eq}}} \right)$ $\therefore \dfrac{1}{{{R}_{plano}}}=\dfrac{2}{{{R}_{eq}}}$ $\therefore {{R}_{plano}}=\dfrac{{{R}_{eq}}}{2}$
Therefore, the radius of curvature of the curved surface of a given plano-convex lens is equal to half of the radius of curvature of an equiconvex lens of focal length $20cm$.
Hence, option A) is correct.
Now, the focus ${{f}_{concave}}$ of a concave mirror made of a silvered plano convex lens is given by
$\dfrac{1}{{{f}_{concave}}}=\dfrac{1}{{{f}_{mirror}}}-\dfrac{2}{{{f}_{lens}}}$ --(4)
Where ${{f}_{mirror}}$ is the focal length of the plane mirror silvered surface while ${{f}_{lens}}$ is the focal length of the convex part of the plano convex lens.
Now, for the given plano convex lens, the focal length of the lens part is given to be $20cm$.
$\therefore {{f}_{lens}}=20cm$. Also, the focal length of a plane mirror is $\infty $.
Therefore, using (4), we get the focal length${{f}_{concave}}$ of the concave mirror formed as a result of the silvered plano convex lens as
$\dfrac{1}{{{f}_{concave}}}=\dfrac{1}{\infty }-\dfrac{2}{20}=0-\dfrac{1}{10}=-\dfrac{1}{10}$ $\left( \because \dfrac{1}{\infty }=0 \right)$ $\therefore {{f}_{concave}}=-10cm$
Therefore, the magnitude of the focal length of the equivalent concave mirror will be $10cm$.
Therefore, option D) is also correct.
Now, according to the lens formula the image distance $v$, object distance $u$ and focal length $f$ of a lens are related as
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ --(5)
Now, let us analyze option B).
For B), the object distance is $u=-15cm$. Image distance is $v=-30cm$. Focal length of the lens is $f=-10cm$
Putting these in values in (5), we get,
$\dfrac{1}{-30}+\dfrac{1}{-15}=\dfrac{1}{-10}$ $\therefore \dfrac{1+2}{30}=\dfrac{1}{10}$ $\therefore \dfrac{3}{30}=\dfrac{1}{10}$
Which is true. Hence, option B) is correct.
Now, let us check option C).
For C), the object distance is $u=-20cm$. Image distance is $v=-40cm$. Focal length of the lens is $f=-10cm$
Putting these in values in (5), we get,
$\dfrac{1}{-40}+\dfrac{1}{-20}=\dfrac{1}{-10}$ $\therefore \dfrac{1+2}{40}=\dfrac{1}{10}$ $\therefore \dfrac{3}{40}=\dfrac{1}{10}$
Which is not true. Hence, option C) is wrong.
Therefore, options A), B) and D) are correct.
Note: This is a complex problem requiring many concepts and formulas. However, there are some common formulas that the student must be completely clear with. The lensmaker’s formula is one of the most important formulas in all of optics. Students should also keep in mind the formula for the focal length of the equivalent concave mirror formed from a silvered plano convex lens since it is a relatively new formula. Students should always be careful and properly apply the sign conventions while plugging in the values in these formulas.