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.

Answer 3

This is a problem involving a concave lens. When a converging beam of light passes through a concave lens, the lens diverges the light, and the focal point shifts based on the lens's focal length. The focal length \( f \) of the concave lens is given as 16 cm, and the beam converges at point P when the lens is placed along the path of the beam. The light initially converges at point P at a distance of 12 cm from the lens (OP = 12 cm). To calculate the new convergence point, we use the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Where:
\( f \) is the focal length of the lens,
\( v \) is the image distance (the distance at which the beam converges after passing through the lens),
\( u \) is the object distance (the initial distance at which the beam converges before passing through the lens).

The object distance \( u \) is -12 cm (since the beam converges at point P, which is 12 cm from the lens). The focal length \( f \) is -16 cm (since it's a concave lens). Substituting the known values into the lens formula: \[ \frac{1}{-16} = \frac{1}{v} - \frac{1}{-12} \] Solving for \( v \): \[ \frac{1}{v} = \frac{1}{-16} + \frac{1}{12} \] \[ \frac{1}{v} = \frac{-3 + 4}{48} = \frac{1}{48} \] \[ v = 48 \, \text{cm} \] Thus, the distance \( x \) from the lens where the beam converges is 48 cm.