Analyzing the Convex Lens Setup

This problem involves two convex lenses, Lens A with a focal length $f_A = 10 \text{ cm}$ and Lens B with a focal length $f_B = 20 \text{ cm}$. They are positioned along the same optical axis with a separation distance '$d$'. We are told that a beam of light, initially parallel to the axis and incident on Lens A, emerges from Lens B also as a parallel beam.

Our objective is to find the value of the separation distance '$d$' in centimeters.

Understanding Convex Lenses and Parallel Light

• A convex lens is a converging lens. When rays of light parallel to the principal axis pass through a convex lens, they converge at a point on the principal axis called the principal focus or focal point ($F$). The distance from the optical center of the lens to the focal point is the focal length ($f$).
• When a parallel beam of light falls on Lens A, it converges towards the focal point $F_A$, located $10 \text{ cm}$ from Lens A.
• For the light beam to emerge as a parallel beam from Lens B, the rays reaching Lens B must be arranged in a specific way relative to Lens B's focal point. Specifically, the rays must either be parallel before hitting Lens B (which isn't the case here, as they converge after Lens A) or they must appear to diverge from Lens B's focal point, or converge towards Lens B's focal point at infinity.
• In this scenario, the rays converge towards $F_A$ after Lens A. This convergence point $F_A$ acts as the object for Lens B. For the rays to become parallel after passing through Lens B, this object must be located at the focal point of Lens B.

Calculating the Lens Separation Distance

Method 1: Using Equivalent Focal Length Formula

For a system of two thin lenses with focal lengths $f_1$ and $f_2$ placed coaxially and separated by a distance $d$, the equivalent focal length ($F_{\text{eq}}$) is given by:

$ \frac{1}{F_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} $

Here, $f_1 = f_A = 10 \text{ cm}$ and $f_2 = f_B = 20 \text{ cm}$.

The condition that a parallel beam enters and a parallel beam leaves means the effective focal length of the lens combination is infinite ($F_{\text{eq}} = \infty$). Therefore, $\frac{1}{F_{\text{eq}}} = 0$.

Substituting this into the formula:

$ 0 = \frac{1}{f_A} + \frac{1}{f_B} - \frac{d}{f_A f_B} $

Plugging in the values for $f_A$ and $f_B$:

$ 0 = \frac{1}{10 \text{ cm}} + \frac{1}{20 \text{ cm}} - \frac{d}{(10 \text{ cm})(20 \text{ cm})} $

First, let's add the focal length terms:

$ \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20 \text{ cm}} $

Now the equation is:

$ 0 = \frac{3}{20 \text{ cm}} - \frac{d}{200 \text{ cm}^2} $

Rearranging to solve for $d$:

$ \frac{d}{200 \text{ cm}^2} = \frac{3}{20 \text{ cm}} $

$ d = \frac{3}{20 \text{ cm}} \times 200 \text{ cm}^2 $

$ d = 3 \times \frac{200}{20} \text{ cm} $

$ d = 3 \times 10 \text{ cm} $

$ d = 30 \text{ cm} $

Method 2: Using Lens Formula and Ray Optics

1. A parallel beam incident on Lens A ($f_A = 10 \text{ cm}$) converges at its focal point $F_A$, which is $10 \text{ cm}$ away from Lens A.

2. These converging rays then travel towards Lens B, placed at a distance '$d$' from Lens A.

3. Let's assume $d > 10 \text{ cm}$. The point $F_A$ is located at a distance of $(d - 10) \text{ cm}$ from Lens B.

4. The rays are converging towards $F_A$. This point $F_A$ serves as the object for Lens B. Since the rays converge towards a point before the lens (in the direction of light travel), this is treated as a virtual object for Lens B. According to the sign convention where light travels from left to right, the object distance $u_B$ is negative. So, $u_B = -(d-10) \text{ cm}$.

5. The problem states that the beam emerges parallel from Lens B. This means the image formed by Lens B is at infinity ($v_B = \infty$).

6. Apply the lens formula for Lens B: $\frac{1}{v_B} - \frac{1}{u_B} = \frac{1}{f_B}$.

7. Substitute the values:

$ \frac{1}{\infty} - \frac{1}{-(d-10 \text{ cm})} = \frac{1}{20 \text{ cm}} $

$ 0 + \frac{1}{d-10 \text{ cm}} = \frac{1}{20 \text{ cm}} $

$ d-10 \text{ cm} = 20 \text{ cm} $

$ d = 30 \text{ cm} $

This result is consistent with the assumption $d > 10 \text{ cm}$.

Conclusion on Distance '$d$'

Both methods confirm that the distance '$d$' between the two convex lenses must be $30 \text{ cm}$ for a parallel beam entering Lens A to emerge as a parallel beam from Lens B.

Reviewing the Options

The options provided are:

• 1. 10
• 2. 20
• 3. 30
• 4. 25
• 5. (empty)

The calculated distance $d = 30 \text{ cm}$ corresponds to option 3.