Physics - Geometrical Optics Question with Solution | TestHub

In convex lens, when an object is placed in between the centre of curvature and focus, the real image is formed behind the centre of curvature. The size of the image is larger than that of the object.

Here, the angle made by inner side of rod with principal axis is $\pi -\frac{2\pi }{3}=\frac{\pi }{3} \mathrm{rad}$.

From figure, height of object is $h_o=2\sin 60°=2\times \frac{\sqrt{3}}{2}=\sqrt{3} \mathrm{cm}$ and given height of image $h_i=\frac{30\sqrt{3}}{13} \mathrm{cm}$

Size of object is $u=\frac{40}{3}+1=\frac{43}{3} \mathrm{cm}$

Now, using lens equation, $\frac{1}{v}-\frac{1}{-u}=\frac{1}{f}$

$\Rightarrow v=\frac{fu}{f-u}$

Image distance, $O{A}^{\text{'}}=\frac{\frac{40}{3}\times 10}{\frac{43}{3}-10}=40 \mathrm{cm}$

And $O{B}^{\text{'}}=\frac{\frac{43}{3}\times 10}{\frac{43}{3}-10}=\frac{430}{13} \mathrm{cm}$

$\therefore A^{\text{'}}{B}^{\text{'}}=40-\frac{430}{13}=\frac{90}{13} \mathrm{cm}$

$\therefore \tan \alpha =\frac{30\sqrt{3}}{13\times \left(\frac{90}{13}\right)}=\frac{1}{\sqrt{3}}$

$\Rightarrow \alpha =\frac{\pi }{6}$

$\therefore n=6.00$