Question

A thin biconvex lens is prepared from the glass (\(\mu = 1.5\)) both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed from the lens at a distance of ________ cm.

• A. 10
• B. 12.5
• C. 13
• D. 13.5

Answer 1

The Correct Option is A

Step 1: Find focal length of the lens Using lens maker's formula: \[ \frac{1}{f_l}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] For a biconvex lens: \[ R_1=+20\,\text{cm}, \qquad R_2=-20\,\text{cm} \] Given \[ \mu=1.5 \] Substituting, \[ \frac{1}{f_l}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right) \] \[ =\frac{1}{2}\left(\frac{1}{20}+\frac{1}{20}\right) \] \[ =\frac{1}{2}\cdot\frac{2}{20} \] \[ =\frac{1}{20} \] Hence, \[ f_l=20\,\text{cm} \] So, power of the lens is \[ P_l=\frac{1}{f_l}=\frac{1}{20}\,\text{cm}^{-1} \] Step 2: Find focal length of silvered surface (mirror) The silvered left surface behaves like a spherical mirror. Radius of curvature: \[ R=20\,\text{cm} \] Focal length of mirror: \[ f_m=\frac{R}{2}=10\,\text{cm} \] Power of mirror: \[ P_m=\frac{1}{f_m}=\frac{1}{10}\,\text{cm}^{-1} \] Step 3: Find equivalent power of lens-mirror system Since light passes through the lens twice and reflects once: \[ P_{eq}=2P_l+P_m \] Substitute values: \[ P_{eq}=2\left(\frac{1}{20}\right)+\frac{1}{10} \] \[ =\frac{1}{10}+\frac{1}{10} \] \[ =\frac{1}{5}\,\text{cm}^{-1} \] Therefore, equivalent focal length is \[ F_{eq}=\frac{1}{P_{eq}}=5\,\text{cm} \] Step 4: Condition for image and object at same position For a mirror system, the image coincides with the object when the object is placed at the center of curvature. So required distance is \[ u=2F_{eq} \] \[ u=2\times5 \] \[ u=10\,\text{cm} \] Final Answer: \[ \boxed{10\,\text{cm}} \]