Question

Assertion (A): For a thin biconvex lens surrounded by air, the minimum distance between a real object and its real image is always equal to two times the focal length of the lens.

Reason (R): The minimum distance between a real object and its real image occurs when object is placed at a distance 2f from the lens.

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is not the correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

Answer: (D)

A is false but R is true.

Answer 2

Analyzing Assertion (A):

The assertion states that for a thin biconvex lens, the minimum distance between a real object and its real image is always equal to two times the focal length ($2f$) of the lens. Let the object distance be $u$ and the image distance be $v$. For a real object, $u$ is negative, and for a real image formed by a convex lens, $v$ is positive.

Let the magnitude of the object distance be $x$, so $u = -x$. The lens formula is:

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

$\frac{1}{v} - \frac{1}{-x} = \frac{1}{f}$

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

$\frac{1}{v} = \frac{1}{f} - \frac{1}{x} = \frac{x-f}{fx}$

$v = \frac{fx}{x-f}$

For a real image, $v > 0$. Since $f > 0$ for a convex lens, we must have $x-f > 0$, which means $x > f$.

The distance between the object and the image, $D$, is the sum of their magnitudes:

$D = |u| + |v| = x + v$

$D = x + \frac{fx}{x-f}$

$D = \frac{x(x-f) + fx}{x-f}$

$D = \frac{x^2 - fx + fx}{x-f}$

$D = \frac{x^2}{x-f}$

To find the minimum distance $D$, we differentiate $D$ with respect to $x$ and set the derivative to zero:

$\frac{dD}{dx} = \frac{d}{dx} \left( \frac{x^2}{x-f} \right)$

Using the quotient rule: $\frac{dD}{dx} = \frac{2x(x-f) - x^2(1)}{(x-f)^2} = \frac{2x^2 - 2fx - x^2}{(x-f)^2} = \frac{x^2 - 2fx}{(x-f)^2}$

Set $\frac{dD}{dx} = 0$:

$\frac{x^2 - 2fx}{(x-f)^2} = 0$

$x^2 - 2fx = 0$

$x(x - 2f) = 0$

Since $x$ (object distance) cannot be zero, we have $x - 2f = 0$, which means $x = 2f$. This value $x=2f$ satisfies the condition $x>f$.

Now, substitute $x=2f$ back into the expression for $D$ to find the minimum distance:

$D_{min} = \frac{(2f)^2}{2f - f} = \frac{4f^2}{f} = 4f$

Therefore, the minimum distance between a real object and its real image is $4f$, not $2f$. So, Assertion (A) is false.

Analyzing Reason (R):

The reason states that the minimum distance between a real object and its real image occurs when the object is placed at a distance $2f$ from the lens. From our derivation above, we found that $x = 2f$ is the condition for the minimum distance $D$. Therefore, Reason (R) is true.

Conclusion: Assertion (A) is false, but Reason (R) is true.