Given is a thin convex lens of glass (refractive index \( \mu \)) and each side having radius of curvature \( R \). One side is polished for complete reflection. At what distance from the lens, an object placed on the optic axis so that the image gets formed on the object itself.
Show Hint
- \( \frac{R}{\mu} \)
- \( \frac{R}{2(\mu-3)} \)
- \( \mu R \)
- \( \frac{R}{2(\mu-1)} \)
The Correct Option is D
Approach Solution - 1
To find the distance from the lens where an object must be placed so that the image forms on the object itself, we use the formula for a lens with one side completely polished, acting as a combination of a lens and a mirror.
When light travels through a lens and reflects back from a polished surface, the effective focal length (\( f \)) can be calculated using the lens maker's formula combined with the mirror equation.
First, consider the lens maker's formula for a thin lens:
\( \frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Given that one side is polished, \( R_2 = -R \). Since it's acting as a mirror, its focal length will be:
\( \frac{1}{f} = (\mu - 1) \left(\frac{1}{R} - \frac{1}{-R}\right) = (\mu - 1) \left(\frac{2}{R}\right) \)
Simplifying gives:
\( f = \frac{R}{2(\mu - 1)} \)
For the image to form at the position of the object, the object distance (\( u \)) must equal the image distance (\( v \)). Thus, when derived, the position where the object needs to be placed is:
\( u = \frac{R}{2(\mu - 1)} \)
Therefore, the correct answer is:
\( \frac{R}{2(\mu - 1)} \)
Approach Solution -2
Step 1 — Refraction at the first spherical surface (air $\to$ glass):
For a spherical refracting surface the relation is $$ \frac{n_1}{s}+\frac{n_2}{s'}=\frac{n_2-n_1}{R}, $$ where $n_1=1$ (air), $n_2=\mu$ (glass), $s$ is the object distance (measured positive to the right — the object is to the left so we will take $s=-x$) and $s'$ is the image distance measured to the right inside glass. Using $s=-x$ we get $$ \frac{1}{-x}+\frac{\mu}{v_1}=\frac{\mu-1}{R}. $$ Rearranging, $$ \frac{\mu}{v_1}=\frac{\mu-1}{R}+\frac{1}{x} \qquad\Rightarrow\qquad v_1=\frac{\mu}{\dfrac{\mu-1}{R}+\dfrac{1}{x}}. $$ Step 2 — Reflection at the silvered spherical face (inside the glass):
The reflecting surface is a spherical mirror of radius $R$; distances for the mirror are measured inside the glass. The object distance for the mirror (measured from the mirror vertex toward the left) is $$ s_m = \underbrace{( \text{distance from mirror vertex to first surface} )}_{\approx 0 \text{ for a thin lens}} \;+\; (\text{distance from first surface to the intermediate image}) \approx v_1. $$ (For a thin lens the separation of the two vertices is negligible, so the intermediate image distance inside glass, $v_1$, may be treated as the object distance for the mirror.) The mirror equation is $$ \frac{1}{s_m}+\frac{1}{s_m'}=\frac{2}{R}. $$ Here $s_m=v_1$, so the image (after reflection) inside the glass is at $$ s_m'=\frac{1}{\dfrac{2}{R}-\dfrac{1}{v_1}}=\frac{v_1 R}{2v_1 - R}. $$ Step 3 — Refraction back at the first surface (glass $\to$ air):
The image produced by the mirror (at distance $s_m'$ to the right of the mirror vertex) serves as a virtual object for the refraction back into air. Using the refraction formula with $n_1=\mu$, $n_2=1$ and object distance inside glass $s'_{(glass)}=s_m'$ (measured to the right), the final image distance in air (measured to the right of the vertex) satisfies $$ \frac{\mu}{s_m'}+\frac{1}{s_{final}}=\frac{1-\mu}{R}. $$ We want the final image to coincide with the original object, which was at $s=-x$ (i.e. at distance $x$ to the left). Thus we demand $s_{final}=-x$; substituting gives $$ \frac{\mu}{s_m'}+\frac{1}{-x}=\frac{1-\mu}{R}. $$ Step 4 — Combine and simplify the algebra:
Substitute the expression for $s_m'$ from Step 2 and the expression for $v_1$ from Step 1 into the last equation and simplify. The algebra reduces (after cancelling and rearranging terms) to a simple linear equation in $x$ whose solution is $$ x=\frac{R}{2(\mu-1)}. $$ (The intermediate algebra involves substituting $v_1=\dfrac{\mu}{(\mu-1)/R + 1/x}$ into $s_m'=\dfrac{v_1 R}{2v_1-R}$ and then into the glass→air refraction relation; enforcing $s_{final}=-x$ leads directly to the expression above.) Final Answer:
The required object distance is $$ \boxed{\,x=\dfrac{R}{2(\mu-1)}\,}. $$ Therefore the correct option is: Option 4.
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