Question

A convex lens of focal length 10 cm form real image of an object. The distance between the object and image may be

• A. 40 cm
• B. 60 cm
• C. Both (1) and (2)
• D. 30 cm

Answer 1

Answer: (C)

Both (1) and (2)

Answer 2

Using the lens formula:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

For $f = 10\, \text{cm}$, we have

$$\frac{1}{10} = \frac{1}{u} + \frac{1}{v}.$$

Multiplying through by $uv$ gives:

$$uv = 10(u+v).$$

Let the distance between the object and the image be $S = u + v$. Then:

$$uv = 10S.$$

The quadratic in $x$ (representing $u$ or $v$) becomes:

$$x^2 - Sx + 10S = 0.$$

For real roots (i.e., real object and image positions), the discriminant must satisfy:

$$\Delta = S^2 - 4 \times 10S \geq 0 \quad \Longrightarrow \quad S^2 - 40S \geq 0.$$

Factorizing:

$$S(S - 40) \geq 0.$$

Since $S > 0$, we must have:

$$S \geq 40\, \text{cm}.$$

Thus, the minimum separation between the object and the image is 40 cm, and any larger separation (like 60 cm) is possible by choosing appropriate object and image distances. Hence, both 40 cm and 60 cm are valid.