Question

A double convex lens whose refractive index is $1.33$ has both radii of curvature of magnitude $10\, cm$. If an object is placed at a distance of $5\, cm$ 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}=10\, cm , R_{2}=-10 \,cm , $
$u=-5 \,cm$ 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=15.15 \,cm$
Now, from lens formula,
$v=\frac{u f}{u+f}=\frac{-5 \times 15.15}{-5+15.15}$
$=\frac{-75.75}{10.15}=-7.46$
$\Rightarrow v$ is -ve hence image will be formed on the same side.