Answer
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Hint: In a Plano-convex lens, one surface is curved outwards, that is, it will be a convex surface and the other surface is found to be flat. The focal length of a Plano-convex lens can be determined using lens maker formula. The focal length of a plane or flat surface is taken to be infinity.
Complete step by step answer:
A lens is a transmissive optical instrument that can focus or disperse an incident light beam by virtue of the phenomenon of refraction of light. A simple lens consists of a single piece of transparent material, while a compound lens consists of several simple lenses or elements usually arranged along a common axis.
Lenses are classified on the basis of the curvature of its two optical surfaces. A lens is said to be biconvex or double convex, if both the surfaces are convex. If both the surfaces have the same radius of curvature, the lens could be called an equi-convex lens. A lens having both surfaces concave is biconcave or double concave lens. If one of the surfaces of the lens is flat, then the lens is called Plano-convex or Plano-concave, depending upon the curvature of the other surface. A lens having one convex and one concave side is called a convex-concave lens.
Plano-convex lens is a type of lens that is plane on one side and convex on the other. Plano-convex lenses have one positive convex face and a flat or plane face on the opposite side of the lens. These lens elements focus parallel light rays into a focal point that is positive and forms a real image that can be projected on a screen or manipulated by spatial filters.
Given that a Plano-convex lens, when silvered at its plane surface is equivalent to a concave mirror of focal length$28cm$, and when its curved surface is silvered and the plane surface is not silvered, it is equivalent to a concave mirror of focal length$10cm$. In the question, we are asked to calculate the refractive index of the material of the lens.
Let ${{f}_{m}}$ be the focal length of the curved silvered surface,
Case 1,
When the convex lens behaves as a concave mirror of focal length$28cm$
Using combinational lens formula, we have,
$\dfrac{1}{f}=\dfrac{2}{{{f}_{1}}}+\dfrac{1}{{{f}_{m}}}$
Where,
$f$is the focal length of the combination (that is, Plano-convex lens)
${{f}_{1}}$is the focal length of convex surface of lens
${{f}_{m}}$is the focal length of mirror surface
We have,
${{f}_{m}}=\infty $, because focal length of a plane surface is infinite
$\dfrac{1}{28}=\dfrac{2}{{{f}_{1}}}+\dfrac{1}{{{f}_{m}}}$
Put ${{f}_{m}}=\infty $
$\begin{align}
& \dfrac{1}{28}=\dfrac{2}{{{f}_{1}}} \\
& {{f}_{1}}=56cm \\
\end{align}$
Case 2,
When the convex lens behaves as a concave mirror of focal length $10cm$
$\begin{align}
& \dfrac{1}{10}=\dfrac{2}{56}+\dfrac{1}{{{f}_{m}}} \\
& \dfrac{1}{{{f}_{m}}}=\dfrac{1}{10}-\dfrac{1}{28} \\
& \dfrac{1}{{{f}_{m}}}=\dfrac{28-10}{280}=\dfrac{18}{280}=\dfrac{9}{140} \\
& {{f}_{m}}=\dfrac{140}{9}cm \\
\end{align}$
The radius of curvature of the lens is,
$\begin{align}
& R=2{{f}_{m}} \\
& R=2\times \dfrac{140}{9}=\dfrac{280}{9}cm \\
\end{align}$
Now, using lens maker formula, we have,
$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
For Plano-convex lens,
${{R}_{2}}=0$
Therefore,
$\dfrac{1}{{{f}_{1}}}=\left( \mu -1 \right)\dfrac{1}{R}$
Substituting the given values,
$\begin{align}
& \mu -1=\dfrac{280}{9}\times \dfrac{1}{56} \\
& \mu =1+\dfrac{5}{9}=\dfrac{14}{9} \\
\end{align}$
Therefore, refractive index of the material of the lens is found to be, $\dfrac{14}{9}$
Hence, the correct option is B.
Note: Plano-convex lens is a combination of two different types of surfaces, one flat and the other curved outwards. The focal length of a plane surface is always infinity. A plane reflecting surface can be considered as a spherical surface of infinite radius of curvature. Since the radius of curvature of a plane surface is infinity, its focal length is also infinity. The focal length of a Plano-convex lens is determined using lens maker formula.
Complete step by step answer:
A lens is a transmissive optical instrument that can focus or disperse an incident light beam by virtue of the phenomenon of refraction of light. A simple lens consists of a single piece of transparent material, while a compound lens consists of several simple lenses or elements usually arranged along a common axis.
Lenses are classified on the basis of the curvature of its two optical surfaces. A lens is said to be biconvex or double convex, if both the surfaces are convex. If both the surfaces have the same radius of curvature, the lens could be called an equi-convex lens. A lens having both surfaces concave is biconcave or double concave lens. If one of the surfaces of the lens is flat, then the lens is called Plano-convex or Plano-concave, depending upon the curvature of the other surface. A lens having one convex and one concave side is called a convex-concave lens.
Plano-convex lens is a type of lens that is plane on one side and convex on the other. Plano-convex lenses have one positive convex face and a flat or plane face on the opposite side of the lens. These lens elements focus parallel light rays into a focal point that is positive and forms a real image that can be projected on a screen or manipulated by spatial filters.
Given that a Plano-convex lens, when silvered at its plane surface is equivalent to a concave mirror of focal length$28cm$, and when its curved surface is silvered and the plane surface is not silvered, it is equivalent to a concave mirror of focal length$10cm$. In the question, we are asked to calculate the refractive index of the material of the lens.
Let ${{f}_{m}}$ be the focal length of the curved silvered surface,
Case 1,
When the convex lens behaves as a concave mirror of focal length$28cm$
Using combinational lens formula, we have,
$\dfrac{1}{f}=\dfrac{2}{{{f}_{1}}}+\dfrac{1}{{{f}_{m}}}$
Where,
$f$is the focal length of the combination (that is, Plano-convex lens)
${{f}_{1}}$is the focal length of convex surface of lens
${{f}_{m}}$is the focal length of mirror surface
We have,
${{f}_{m}}=\infty $, because focal length of a plane surface is infinite
$\dfrac{1}{28}=\dfrac{2}{{{f}_{1}}}+\dfrac{1}{{{f}_{m}}}$
Put ${{f}_{m}}=\infty $
$\begin{align}
& \dfrac{1}{28}=\dfrac{2}{{{f}_{1}}} \\
& {{f}_{1}}=56cm \\
\end{align}$
Case 2,
When the convex lens behaves as a concave mirror of focal length $10cm$
$\begin{align}
& \dfrac{1}{10}=\dfrac{2}{56}+\dfrac{1}{{{f}_{m}}} \\
& \dfrac{1}{{{f}_{m}}}=\dfrac{1}{10}-\dfrac{1}{28} \\
& \dfrac{1}{{{f}_{m}}}=\dfrac{28-10}{280}=\dfrac{18}{280}=\dfrac{9}{140} \\
& {{f}_{m}}=\dfrac{140}{9}cm \\
\end{align}$
The radius of curvature of the lens is,
$\begin{align}
& R=2{{f}_{m}} \\
& R=2\times \dfrac{140}{9}=\dfrac{280}{9}cm \\
\end{align}$
Now, using lens maker formula, we have,
$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
For Plano-convex lens,
${{R}_{2}}=0$
Therefore,
$\dfrac{1}{{{f}_{1}}}=\left( \mu -1 \right)\dfrac{1}{R}$
Substituting the given values,
$\begin{align}
& \mu -1=\dfrac{280}{9}\times \dfrac{1}{56} \\
& \mu =1+\dfrac{5}{9}=\dfrac{14}{9} \\
\end{align}$
Therefore, refractive index of the material of the lens is found to be, $\dfrac{14}{9}$
Hence, the correct option is B.
Note: Plano-convex lens is a combination of two different types of surfaces, one flat and the other curved outwards. The focal length of a plane surface is always infinity. A plane reflecting surface can be considered as a spherical surface of infinite radius of curvature. Since the radius of curvature of a plane surface is infinity, its focal length is also infinity. The focal length of a Plano-convex lens is determined using lens maker formula.
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