Two convex lenses of focal length 20 cm each are placed coaxially with a separation of 60 cm between them. The image of the distant object formed by the combination is at cm from the first lens.
Two convex lenses of focal length 20 cm each are placed coaxially with a separation of 60 cm between them. The image of the distant object formed by the combination is at cm from the first lens.
Correct Answer: 100
Solution and Explanation
Solution:
Given:
Focal length of each lens, \( f_1 = f_2 = 20 \, \text{cm} \)
Separation between the lenses, \( d = 60 \, \text{cm} \)
The object is at a considerable distance (assumed to be at infinity).
Objective: Find the distance of the final image from the first lens.
Approach: Step 1: Image Formation by the First Lens For the first lens (\( f_1 = 20 \, \text{cm} \)), the object is at infinity. Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where:
\( u \) is the object distance (infinity),
\( v \) is the image distance,
\( f \) is the focal length.
Plugging in the values: \[ \frac{1}{20} = \frac{1}{v} - \frac{1}{\infty} \implies \frac{1}{v} = \frac{1}{20} \implies v = 20 \, \text{cm} \] So, the first lens forms an image at 20 cm from itself.
Step 2: Object for the Second Lens The image formed by the first lens acts as the object for the second lens. The separation between the lenses is 60 cm, and the first image is 20 cm from the first lens. Therefore, the distance of this image (object for the second lens) from the second lens is: \[ u_2 = d - v_1 = 60 \, \text{cm} - 20 \, \text{cm} = 40 \, \text{cm} \] Since the image is on the same side as the incoming light for the second lens, we consider \( u_2 \) as negative in the lens formula (real image for the first lens acts as a virtual object for the second lens).
Step 3: Image Formation by the Second Lens Using the lens formula for the second lens (\( f_2 = 20 \, \text{cm} \)): \[ \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \] Plugging in the values: \[ \frac{1}{20} = \frac{1}{v_2} - \frac{1}{-40} \implies \frac{1}{20} = \frac{1}{v_2} + \frac{1}{40} \] Solving for \( v_2 \): \[ \frac{1}{v_2} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40} = \frac{1}{40} \implies v_2 = 40 \, \text{cm} \] The positive sign indicates that the final image is formed on the opposite side of the second lens from where the light is coming.
Step 4: Distance of the Final Image from the First Lens The final image is 40 cm from the second lens. Since the lenses are 60 cm apart, the distance from the first lens is: \[ \text{Total distance} = d + v_2 = 60 \, \text{cm} + 40 \, \text{cm} = 100 \, \text{cm} \]
Conclusion: The image of the distant object formed by the combination of the two convex lenses is located at \boxed{100 \, \text{cm}} from the first lens.
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