Physics - Geometrical Optics Question with Solution | TestHub

$$\frac{1}{f_{lens}}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

Substituting the given values:

$$\frac{1}{f_{lens}}=(0.5)\left(\frac{1}{25}+\frac{1}{25}\right)$$

$$\frac{1}{f_{lens}}=(0.5)\left(\frac{2}{25}\right)$$

So, $f_{lens}=25$ cm.

Substituting these values into the equivalent focal length formula:

$$-\frac{1}{F_{eq}}=\frac{2}{25}+\frac{1.5}{25/2}$$

$$\frac{1}{F_{eq}}=-\frac{4}{25}$$

For the image to be formed on the object itself, the object must be placed at the center of curvature of the equivalent mirror. The center of curvature for a mirror is at a distance equal to twice its focal length.

Therefore, the object distance $l$ is:

$l=2\times F_{eq}$

The question asks for the value of $\frac{2l}{5}$.

$$\frac{2l}{5}=\frac{2\times 12.5}{5}$$

$$\frac{2l}{5}=\frac{25}{5}$$

$$\frac{2l}{5}=5$$

The final answer is $\boxed{5}$.