
When the plane surface of a plano-convex lens of refractive index \[1.5\] is silvered, it behaves like a concave mirror with $f=30cm$. When its convex surface is silvered, it will behave like a concave mirror of focal length:
\[\begin{align}
& A.10cm \\
& B.20cm \\
& C.30cm \\
& D.45cm \\
\end{align}\]
Answer
580.2k+ views
Hint: Here we have a plano-convex lens, which is being silvered on one side, that is either the plane side or the convex side is silvered. Clearly, if the lens is silvered, it acts as a mirror. Here, it is given that it will act as a concave mirror. Using the lens formula we can find focal length.
Formula used: $\dfrac{1}{f_{C}}=(\mu-1)\dfrac{1}{R}$
Complete step by step answer:
It is given that the refractive index plano-convex lens is \[1.5\].
First, when the plane side is silvered, we get a mirror with focal length $f=30cm$
Then the light ray gets refracted into the lens, gets reflected at the plane side and gets refracted again outside the lens.
We know that the power of plane mirror is $P_{P}=0$ and the power of the curve surface be $P_{C}$
Then the total power of the system is $P=2P_{C}-P_{P}=2P_{C}$
We know $\dfrac{1}{f}=P$
Then, $\dfrac{2}{f_{C}}=\dfrac{1}{30}$ where, $f_{C}$ is the focal length of the lens.
$\implies f_{C}=60$
From lens makers formula we know that $\dfrac{1}{f_{C}}=(\mu-1)\dfrac{1}{R}$ where $R$ is the radius of curvature of the lens. Substituting we get, $\dfrac{1}{60}=(1.5-1)\dfrac{1}{R}$
$\implies R=30cm$
The radius of the convex surface is \[30cm\].
Similarly, if the convex surface is silvered, then the $P=2P_{C}-P_{M}=-P_{M}$, here $P_{C}=0$ as it is silvered.
Then the $\dfrac{1}{f_{M}}= -P_{M}$
Since the radius of the curvature of the plano-convex lens remains the same, we can say that $f_{M}=R=30$
So, the correct answer is “Option C”.
Note: Thus for a plano-spherical mirror, the focal length is equal to the radius of the curvature of the mirror. Plano-spherical mirrors are used in adventure parks, as they produce different sizes of the images. Also a lens can be converted to a mirror by silvering one of its sides.
Formula used: $\dfrac{1}{f_{C}}=(\mu-1)\dfrac{1}{R}$
Complete step by step answer:
It is given that the refractive index plano-convex lens is \[1.5\].
First, when the plane side is silvered, we get a mirror with focal length $f=30cm$
Then the light ray gets refracted into the lens, gets reflected at the plane side and gets refracted again outside the lens.
We know that the power of plane mirror is $P_{P}=0$ and the power of the curve surface be $P_{C}$
Then the total power of the system is $P=2P_{C}-P_{P}=2P_{C}$
We know $\dfrac{1}{f}=P$
Then, $\dfrac{2}{f_{C}}=\dfrac{1}{30}$ where, $f_{C}$ is the focal length of the lens.
$\implies f_{C}=60$
From lens makers formula we know that $\dfrac{1}{f_{C}}=(\mu-1)\dfrac{1}{R}$ where $R$ is the radius of curvature of the lens. Substituting we get, $\dfrac{1}{60}=(1.5-1)\dfrac{1}{R}$
$\implies R=30cm$
The radius of the convex surface is \[30cm\].
Similarly, if the convex surface is silvered, then the $P=2P_{C}-P_{M}=-P_{M}$, here $P_{C}=0$ as it is silvered.
Then the $\dfrac{1}{f_{M}}= -P_{M}$
Since the radius of the curvature of the plano-convex lens remains the same, we can say that $f_{M}=R=30$
So, the correct answer is “Option C”.
Note: Thus for a plano-spherical mirror, the focal length is equal to the radius of the curvature of the mirror. Plano-spherical mirrors are used in adventure parks, as they produce different sizes of the images. Also a lens can be converted to a mirror by silvering one of its sides.
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