Question

A biconvex lens having radius of curvature \(20\,\text{cm}\) for both surfaces and one side of the lens is silvered as shown in the figure. Object is at distance \(x\) cm from the lens. Find \(x\) such that image is on the object itself. Given \( \mu = 1.5 \).

Answer 1

Correct Answer: 10

Step 1: Understanding the Concept:
A silvered lens behaves as a concave mirror.
To form the image at the position of the object, the object must be placed at the center of curvature of this equivalent mirror.
Step 2: Key Formula or Approach:
Focal length of lens is given by $\frac{1}{f_L} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Focal length of mirror is $f_m = \frac{-R}{2}$.
Effective focal length of the combination is $\frac{1}{F} = \frac{2}{f_L} + \frac{1}{f_m}$.
Step 3: Detailed Explanation:
For the biconvex lens: $R_1 = 20 \text{ cm}, R_2 = -20 \text{ cm}, \mu = 1.5$.
\[ \frac{1}{f_L} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} \right) = 0.5 \times \frac{2}{20} = \frac{1}{20} \implies f_L = 20 \text{ cm} \]
Focal length of the silvered surface (acting as mirror): $f_m = \frac{-20}{2} = -10 \text{ cm}$.
Using the power of combination formula:
\[ \frac{-1}{F} = \frac{2}{f_L} + \frac{1}{f_m} \]
Wait, the standard sign convention formula for equivalent mirror is $\frac{1}{F} = \frac{2}{f_L} - \frac{1}{f_m}$ depending on convention. Using standard power $P = 2P_L + P_m$:
\[ P_L = \frac{1}{f_L} = \frac{1}{20} \]
\[ P_m = -\frac{1}{f_m} = -\frac{1}{-10} = \frac{1}{10} \]
\[ P_{eq} = 2\left(\frac{1}{20}\right) + \frac{1}{10} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \]
Since $P_{eq} = -\frac{1}{F}$, we have $F = -5 \text{ cm}$.
The system acts as a concave mirror of focal length 5 cm.
To coincide image with object, distance $u = 2f = 2 \times 5 = 10 \text{ cm}$.
So, $x = 10 \text{ cm}$.
Step 4: Final Answer:
The value of $x$ is 10 cm.