Question

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.

• A. \( \frac{R}{\mu} \)
• B. \( \frac{R}{2(\mu-3)} \)
• C. \( \mu R \)
• D. \( \frac{R}{2(\mu-1)} \)

Hint:

For a convex lens with one polished surface, the focal length can be calculated by considering both surfaces and using the lens maker's formula.

Answer 1

The Correct Option is D

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:

\( \frac{R}{2(\mu - 1)} \)

Answer 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.