© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down known quantities and concepts
• Refractive index of the plano-convex lens, $ \mu = 1.5 $
• Radius of curvature of the curved surface, $ R = 30 \text{ cm} $
• Plano side is taken as $ R_1 = \infty $ and the convex side as $ R_2 = -30 \text{ cm} $ (negative because it is curved outward from the lens).
• The lens is silvered at the curved surface, effectively creating a lens–mirror system.
Step 2: Calculate the focal length of the lens ($ f_\ell $)
For a thin lens in air, the lens formula is:
$ \dfrac{1}{f_\ell} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) $
Plug in the numerical values:
$ \dfrac{1}{f_\ell} = (1.5 - 1)\left(\dfrac{1}{\infty} - \dfrac{1}{-30}\right) = 0.5 \left( 0 + \dfrac{1}{30} \right) = \dfrac{0.5}{30} = \dfrac{1}{60}. $
Thus,
$ f_\ell = 60 \text{ cm}. $
Step 3: Express the effective focal length of the lens–mirror combination ($ F $)
When the curved surface of the lens is silvered, the system can be treated as a mirror whose power is given by:
$ \dfrac{1}{F} = 2 \left(\dfrac{1}{f_\ell}\right) + \dfrac{1}{f_m}, $
where
• $ \dfrac{1}{f_\ell} = \dfrac{1}{60} $, from above, and
• $ f_m $ is the mirror’s focal length for reflection from the plane side. Since the curvature radius is 30 cm for the lens, the mirror formed effectively has $ f_m = \dfrac{R}{2} = \dfrac{30}{2} = 15 \text{ cm} $ for reflection from that side.
So,
$ \dfrac{1}{F} = 2 \left(\dfrac{1}{60}\right) + \dfrac{1}{15}. $
Compute each term:
$ 2 \left(\dfrac{1}{60}\right) = \dfrac{2}{60} = \dfrac{1}{30}, \quad \dfrac{1}{15} = \dfrac{2}{30}. $
Hence,
$ \dfrac{1}{F} = \dfrac{1}{30} + \dfrac{2}{30} = \dfrac{3}{30} = \dfrac{1}{10} \quad \Longrightarrow \quad F = 10 \text{ cm}. $
Step 4: Use the condition for an image of the same size
For a mirror (or a mirror-like system) to produce a real image of the same size as the object, the object must be placed at $ 2F $ (the center of curvature position):
$ u = 2F = 2 \times 10 = 20 \text{ cm}. $
Thus, the object must be placed 20 cm from the lens–mirror combination to get a real image of the same size as the object.
Final Answer
The correct distance at which the object should be placed is $20 \text{ cm}.$
