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. |
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