question_answer

There is a hole in centre of a thin circular biconvex lens of focal length 4 cm. Diameter of hole is half of diameter of the lens. A point source of light is placed 9 cm from the wall and lens is placed in between, such that a single circular illuminated spot with sharp bright edges is formed on the wall. Distance of lens and source must be

A) 1 cm

B) 3 cm

C) 5 cm    

D) 4 cm

Correct Answer: B

Solution :

To form a sharp image, the light cone that passes the circular hole and light cone that is formed by the refracted light have same base area on screen. This is shown below.
Now, from lens equation, we have
\[\frac{1}{v}-\frac{1}{u}-\frac{1}{f}\]\[\frac{1}{v}=\frac{1}{f}+\frac{1}{-x}=\frac{x-f}{fx}\]
Image distance\[=\,v\,=\frac{fx}{x-f}=\frac{fu}{u-f}\]
Let, h = radius of image on screen,
r = radius of hole in lens and R = radius of lens.
Now, from similar triangles \[\Delta OAB\] and\[\Delta OCD,\]we have
\[\frac{h}{r}=\frac{d}{u}\]	?.. (i)
Note That no sign convention is required here. And from similar triangles \[\Delta DCI\] and \[\Delta AEI\], we have

\[\frac{h}{R}=\frac{u+v-d}{v}\left( \therefore R=2r\,and\,v=\frac{fu}{u-f} \right)\]
\[\Rightarrow \,\,\,\,\frac{h}{2r}=\frac{u+\left( \frac{fu}{u-f} \right)-d}{\left( \frac{fu}{u-f} \right)}\]	?.. (ii)
From Eqs. (i) and (ii) ,we have
\[\Rightarrow \,\,\,\,\frac{d}{2u}=\frac{u\left( u-f \right)+fu-d\left( u-f \right)}{fu}\]\[\Rightarrow \,\,\,\,2{{u}^{2}}-2du+df=0\]\[\Rightarrow \,\,\,\,u=\frac{2d\pm \sqrt{4{{d}^{2}}-8df}}{4}\]\[\Rightarrow \,\,\,\,u=\frac{d}{2}\pm \left( \frac{1}{2}\sqrt{d\left( d-2f \right)} \right)\]
Here, \[f=4cm,\text{ }d=9cm\] So, \[u=\frac{9}{2}\pm \frac{1}{2}\sqrt{9\left( 9-8 \right)}=\frac{9}{2}\pm \frac{3}{2}\]
\[\therefore \,\,\,u=6cm\] or \[u=3cm\]
So, both are possible postions.