Question

The following figure shows a beam of light converging at point \(P\). When a concave lens of focal length \(16 \,cm\) is introduced in the path of the beam at a place shown by dotted line such that \(OP\) becomes the axis of the lens, the beam converges at a distance \(x\) from the lens. The value of \(x\) will be equal to

• A. 12 cm
• B. 24 cm
• C. 36 cm
• D. 48 cm

Answer 1

The Correct Option is D

So, here when we put the concave lens,
let the beam will converge at a distance \(x=v\)
Using lens formulae, we have, \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
Where \(u =12\, cm\) and \(f =-16\, cm\) is given
\(\therefore \frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)
\(\left( -\frac{1}{16} \right) + \left( \frac{1}{12} \right) = \frac{1}{48} \, \text{cm}\)
\(\Rightarrow v =48 \,cm\)
Hence, \(x=48 \,cm\)

Answer 2

For a concave lens, the focal length \(f\) is negative. The formula for the image distance \(v\) formed by a lens is given by the lens equation: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] Where:
\(f\) is the focal length of the lens,
\(u\) is the object distance (distance from the lens to the point where the light beam converges),
\(v\) is the image distance (the distance at which the beam will converge after passing through the lens).
In this case, \(f = -16\) cm (since the lens is concave), and the object distance \(u = -12\) cm (since the beam converges at point P).
Substituting into the lens formula: \[ \frac{1}{-16} = \frac{1}{-12} + \frac{1}{v} \] Solving for \(v\), we get: \[ v = 48 \, \text{cm} \] Thus, the beam will converge at a distance of 48 cm from the lens.