Question

Two thin convex lenses L₁ and L₂,have focal lengths 4 cm and 10 cm, respectively. They are separated by a distance of x cm as shown in the figure. A point source S is placed on the principal axis at a distance 12 cm to the left of L₁. If the image of S is formed at infinity, the value of x is:

• A. 6
• B. 16
• C. 14
• D. 24
• E. 10

Answer 1

The Correct Option is B

Given:

• Focal length of lens \( L_1 \), \( f_1 = 4 \, \text{cm} \) (convex)
• Focal length of lens \( L_2 \), \( f_2 = 10 \, \text{cm} \) (convex)
• Distance between \( L_1 \) and \( L_2 \): \( x \, \text{cm} \)
• Point source \( S \) is placed 12 cm to the left of \( L_1 \).
• Final image is formed at infinity.

Step 1: Image Formation by \( L_1 \)

For \( L_1 \), using the lens formula:

\[ \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \]

Here, \( u_1 = -12 \, \text{cm} \) (object distance is negative by convention), \( f_1 = 4 \, \text{cm} \):

\[ \frac{1}{v_1} - \frac{1}{-12} = \frac{1}{4} \]

\[ \frac{1}{v_1} = \frac{1}{4} - \frac{1}{12} = \frac{3 - 1}{12} = \frac{2}{12} = \frac{1}{6} \]

Thus, the image distance \( v_1 = 6 \, \text{cm} \) (to the right of \( L_1 \)).

Step 2: Condition for Final Image at Infinity

For the final image to be at infinity, the image formed by \( L_1 \) must lie at the focal point of \( L_2 \). Therefore:

\[ x - v_1 = f_2 \]

\[ x - 6 = 10 \]

\[ x = 16 \, \text{cm} \]

Conclusion:

The required separation \( x \) between the lenses is 16 cm.

Answer: \(\boxed{B}\)

Answer 2

Let $f_1$ be the focal length of lens L₁ and $f_2$ be the focal length of lens L₂. 

Given $f_1 = 4$ cm and $f_2 = 10$ cm. Let $u_1$ be the object distance for L₁. $u_1 = -12$ cm (since it is to the left of L₁). 

Let $v_1$ be the image distance for L₁. Using the lens formula for L₁: $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$ $\frac{1}{v_1} - \frac{1}{-12} = \frac{1}{4}$ $\frac{1}{v_1} = \frac{1}{4} - \frac{1}{12}$ $\frac{1}{v_1} = \frac{3-1}{12} = \frac{2}{12} = \frac{1}{6}$ $v_1 = 6$ cm. 

The image formed by L₁ acts as an object for L₂. Let $u_2$ be the object distance for L₂. 

The separation between the lenses is $x$ cm. 

The image formed by L₁ is at a distance of 6 cm to the right of L₁. $u_2 = -(x-6)$ cm (negative as the object is to the right of L₂). The image is formed at infinity, so $v_2 = \infty$. 

Using the lens formula for L₂: $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$ $\frac{1}{\infty} - \frac{1}{-(x-6)} = \frac{1}{10}$ $0 + \frac{1}{x-6} = \frac{1}{10}$ $x-6 = 10$ $x = 10+6$ $x = 16$ cm. 

Final Answer: The final answer is $\boxed{B}$