Question: A converging beam of light converges at a point $P$. When a diverging lens of focal length $f$ is pl...

Question

A converging beam of light converges at a point $P$ . When a diverging lens of focal length $f$ is placed in the path of the converging beam at a distance 10 cm from $P$ , the beam converges at a distance 2 cm from $P$ .

The focal length of the lens is

Answer 1

Solution

1. Understanding the setup: A converging beam of light means that the rays are heading towards a specific point. If a lens is placed in the path of this beam, the point where the rays would have converged acts as a virtual object for the lens.

2. Identifying the object: The beam would converge at point P. This point P acts as a virtual object for the diverging lens.

3. Determining the object distance (u): The lens is placed at a distance of 10 cm from P. Since P is a virtual object (the rays are converging towards it), according to the Cartesian sign convention (light travels from left to right, distances measured to the right of the lens are positive), the object distance u is positive. So, u = +10 cm.

4. Determining the image distance (v): The problem states that after passing through the diverging lens, the beam converges at a distance 2 cm from P. A diverging lens (concave lens) causes light rays to spread out more or converge less.

If the rays were converging to P, a diverging lens will push the convergence point further away from the lens, or even make them diverge completely.
Since the beam still converges, the new convergence point (let's call it P') must be further from the lens than P.
- Distance from lens to P = 10 cm.
- Distance from lens to P' = Distance from lens to P + 2 cm = 10 cm + 2 cm = 12 cm.
- Since P' is on the side where the light exits the lens (and is a real convergence point), the image distance v is positive.

So, v = +12 cm.

(Note: If we considered P' to be 2 cm to the left of P, then v = 10 - 2 = +8 cm. Applying the lens formula would yield f = +40 cm, which corresponds to a converging lens, contradicting the problem statement that it's a diverging lens. Hence, P' must be further away.)

5. Applying the lens formula: The lens formula is given by: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Substitute the values of u and v: $\frac{1}{+12} - \frac{1}{+10} = \frac{1}{f}$ $\frac{1}{12} - \frac{1}{10} = \frac{1}{f}$

To subtract the fractions, find a common denominator, which is 60: $(\frac{5}{60}) - (\frac{6}{60}) = \frac{1}{f}$ $-\frac{1}{60} = \frac{1}{f}$

6. Calculating the focal length (f): $f = -60 cm$

The negative sign confirms that it is a diverging lens, which is consistent with the problem statement.