Question

A double convex lens whose refractive index is $1.33$ has both radii of curvature of magnitude $10cm$. If an object is placed at a distance of $5cm$ from this lens, the position of the image formed is

• A. $7.46$ same side of the object
• B. $7.46$ opposite side of the object
• C. $14.45$ same side of the object
• D. $14.45$ opposite side of the object

Answer 1

Answer: (A)

Using $\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
here, $R_1=10cm,R_2=-10cm,$
$u=-5cm$ and $\mu =1.33$
$\frac{1}{f}=(1.33-1)\left(\frac{1}{10}+\frac{1}{10}\right)$
$\frac{1}{f}=0.33\times \frac{2}{10}=\frac{0.33}{5}$
$\Rightarrow f=15.15cm$
Now, from lens formula,
$v=\frac{uf}{u+f}=\frac{-5\times 15.15}{-5+15.15}$
$=\frac{-75.75}{10.15}=-7.46$
$\Rightarrow v$ is -ve hence image will be formed on the same side.