question_answer

A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

A) 0.55 cm towards the lens

B) 0

C) 0.55 cm away from the lens

D) 1.1 cm away from the lens

Correct Answer: C

Solution :

\[U\,\,\,=\,\,-10\] \[V\,\,=\,\,10\] \[f\text{ }=\text{ }5\text{ }cm\] Shift due to slab = \[t\left( 1-\frac{1}{\mu } \right)\] \[=\,\,\,\,1.5\left( 1-\frac{1}{1.5} \right)\] \[=\,\,\,\,1.5\,\,\times \,\,\frac{1}{3}\,\,=\,\,\,0.5\,cm\] Now for lens \[u\,\,=\,\,-\,9.5\,cm\]                        \[f\,\,=\,\,5\,cm\] \[\frac{1}{v}\,\,-\,\,\frac{1}{u}\,\,=\,\,\frac{1}{f}\] \[\frac{1}{v}\,\,-\,\,\frac{1}{-9.5}\,\,=\,\,\frac{1}{5}\] \[\frac{1}{v}\,\,=\,\,\frac{1}{5}\,\,=\,\,\frac{1}{9.5}\] \[v\,\,=\,\,\frac{5\,\,\times \,\,9.5}{4.5}\] \[v=\text{ }10.55\text{ }cm\] So, screen is shifted away from lens by 0.55 cm.