Final Image Position for Biconvex Lens and Plane Mirror Combination

To find the nature and position of the final image, we consider the optical combination in three sequential steps:

Step 1: First Refraction through the Biconvex Lens

• Focal length of the biconvex lens, $f = +15\text{ cm}$
• Object distance, $u_1 = -30\text{ cm}$ (to the left of the lens)

Using the thin lens formula:
$\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f}$

Substitute the given values:
$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{15}$

$\frac{1}{v_1} = \frac{1}{15} - \frac{1}{30} = \frac{1}{30}$

$v_1 = +30\text{ cm}$

Thus, the first image $I_1$ would be formed at a distance of $30\text{ cm}$ to the right of the lens.

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Step 2: Reflection from the Plane Mirror

• The plane mirror is placed $10\text{ cm}$ to the right of the lens.
• Light rays are intercepted by the mirror before reaching $I_1$.
• Therefore, $I_1$ acts as a virtual object for the plane mirror.

Distance of virtual object $I_1$ behind the plane mirror:
$u_m = 30\text{ cm} - 10\text{ cm} = +20\text{ cm}$

A plane mirror forms an image at an equal distance in front of it for a virtual object. Hence, the reflected rays converge to form a real image $I_2$ at a distance of $20\text{ cm}$ in front of the mirror (to the left of the mirror).

Since the mirror is $10\text{ cm}$ to the right of the lens, $I_2$ is located:
$20\text{ cm} - 10\text{ cm} = 10\text{ cm}\text{ to the left of the lens}$

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Step 3: Second Refraction through the Lens

The reflected rays travel from right to left (toward the lens). Since they are converging toward $I_2$ (which lies behind the lens relative to the direction of propagation of the reflected light):

• $I_2$ acts as a virtual object for the lens.
• Taking the direction of propagation of light (right to left) as positive:
  • Object distance, $u_2 = +10\text{ cm}$
  • Focal length of the lens, $f = +15\text{ cm}$

Using the lens formula again:
$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f}$

$\frac{1}{v_2} - \frac{1}{10} = \frac{1}{15}$

$\frac{1}{v_2} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$

$v_2 = +6\text{ cm}$

Since $v_2 > 0$, the final image $I_3$ is real and is formed at a distance of $6\text{ cm}$ to the left of the lens.

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Final Position Relative to the Mirror:

The distance of the final image from the plane mirror is:
$\text{Distance} = \text{Distance between Lens and Mirror} + \text{Distance of } I_3 \text{ from Lens}$
$\text{Distance} = 10\text{ cm} + 6\text{ cm} = 16\text{ cm}$

Hence, the final image is real and at a distance of $16\text{ cm}$ from the mirror.

Correct Answer: B