Question

A point object is moving at a constant speed of 1 ms^-1 along the principal axis of a convex lens of focal length 10cm. The speed of the image is also 1 ms^-1 , when the object is at _______ cm from the optic centre of the lens.

• A. 15
• B. 20
• C. 5
• D. 10

Answer 1

The Correct Option is B

Given Information:
Speed of object, \( v_o = 1\,ms^{-1} \)
Speed of image, \( v_i = 1\,ms^{-1} \)
Focal length of convex lens, \( f = 10\,cm \)

Step-by-Step Explanation:

Step 1: Relation between object speed and image speed

Using lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where:
\( u \) = object distance (negative for real objects),
\( v \) = image distance (positive for real images).

Step 2: Differentiate lens formula with respect to time to find relation between velocities

Differentiating both sides: \[ 0 = -\frac{1}{v^2}\frac{dv}{dt} + \frac{1}{u^2}\frac{du}{dt} \]

Thus, \[ \frac{dv}{dt} = \frac{v^2}{u^2}\frac{du}{dt} \]

Given, \( |\frac{dv}{dt}| = |\frac{du}{dt}| \), thus: \[ \frac{v^2}{u^2} = 1 \quad\Rightarrow\quad v^2 = u^2 \]

Thus, \( v = u \) (since distances are positive for simplicity).

Step 3: Use \( v = u \) in lens formula to find object distance

From lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}, \quad\text{and since}\quad v = u: \]

\[ \frac{1}{f} = \frac{1}{u} - \frac{1}{u} = 0 \]

But this leads to a contradiction if we directly substitute \( v = u \). Hence, let's carefully reconsider signs:

For a real object, \( u \) is negative and \( v \) is positive. Therefore, the correct condition considering signs becomes: \[ v = -u \]

Substitute \( v = -u \) into lens formula carefully: \[ \frac{1}{f} = \frac{1}{-u} - \frac{1}{u} = -\frac{2}{u} \]

Thus: \[ -\frac{2}{u} = \frac{1}{f} \quad\Rightarrow\quad u = -2f \]

Since \( f = 10\,cm \), object distance is: \[ u = -2 \times 10\,cm = -20\,cm \]

Thus, the object must be at \(20\,cm\) from the lens.