question_answer

A slab of a material of refractive index 2 is shown in figure  has a curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in the figure. An object O is placed at a distance of 15 cm from the pole P as shown. The distance of the final image of O from P, as viewed from the left is

A) 10 cm              

B) 20 cm

C) -30 cm             

D) -20 cm

Correct Answer: C

Solution :

[c] : When light travels from medium of refractive index \[{{\mu }_{2}}\]to medium of refractive index \[{{\mu }_{1}}\]at a single spherical surface, the formula used is \[\frac{{{\mu }_{1}}}{v}-\frac{{{\mu }_{2}}}{u}=\frac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\] Direction of light is in positive direction. \[\therefore \]\[\frac{1.0}{v}-\frac{2.0}{-15}=\frac{1.0-2.0}{-10}\] F is centre of curvature of AP5. or\[\frac{1}{v}=\frac{1}{10}-\frac{2}{15}=\frac{3-4}{30}=\frac{-1}{30}\]or\[v=-30cm\] \[\therefore \]The distance of the final image of 0 from P, as viewed from the left, is 30 cm to right of P. The image formed will be virtual at E.