Discussion Forum

1)

 A half-spherical glass lens with a refractive index 1.5 is placed in a liquid with a refractive index of 1.3 (see the following figure) . The radius of the half spherical lens is 10 cm. A parallel  beam of light travelling  in the liquid  is refracted by the glass  lens, Then the absolute  value of the position  of the  image from the centre of the glass lens will be 

Answer:

Option B

Explanation:

 Focal length of the lines

 $\frac{1}{f}=\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

 Here, refractive index for galss lens, $n_{2}=1.5$

 Refractive index for water , $n_{1}=1.3$

 Radius of curvature  , $R_{1}=10 cm$

                                  $R_{2}= \infty$

 Putting these values , we get

  $\frac{1}{f}=\left(\frac{1.5}{1.3}-1\right)\left(\frac{1}{10}-\frac{1}{\infty}\right)$

 $\frac{1}{f}=\frac{0.15}{10}$

 $\Rightarrow$    f=65 cm

 Now parallel beam of light will converge  at the focus of lens, so image position v=65 cm.