Question

A concave lens of radius of curvature 15 cm and \(\mu\) = 1.5 is placed in water \(\left (\mu \dfrac{4}{3} \right )\). If one surface is silvered, then image distance from lens when an object is placed at distance of 14 cm from the lens is


Object

• A. 10.5 cm
• B. 4.2 cm
• C. 8.5 cm
• D. 6.4 cm

Answer 1

The correct option is B 4.2 cm

$\frac{1}{f_{lens}}=(\frac{1.5}{4/3}-1)\left(\frac{-2}{15}\right)\Rightarrow flens=-60cm\therefore \frac{1}{f_e}=\frac{2}{-60}+\frac{2}{-15}\Rightarrow f_e=-6cm\therefore V=\frac{14\times (-6)}{14+6}=4.2cm\text{from lens (virtual)}$