Q47. What would be the position of image formed by the lens combination given in the figure below?
f

Q: What is the correct answer for CUET UG Physics Q47?

The correct answer is a. This problem requires sequential application of the lens formula, 1/f = 1/v - 1/u, where f is focal length, u is object distance, and v is image distance. Distances to the left of the lens are negative, to the right are positive. Lenses are at x=0, x=10cm, x=20cm. 1. For the first lens (Convex, f₁ = +10 cm), object is at u₁ = -20 cm. 1/+10 = 1/v₁ - 1/(-20) => 1/v₁ = 1/10 - 1/20 = 1/20. So, v₁ =

Q: Q47. What would be the position of image formed by the lens combination given in the figure below? f=+10cm 0 20 cm -10 cm 10 cm 20 cm +10 cm (a) 10 cm to the right of third lens (b) 7 cm to the right…

The correct answer is a. This problem requires sequential application of the lens formula, 1/f = 1/v - 1/u, where f is focal length, u is object distance, and v is image distance. Distances to the left of the lens are negative, to the right are positive. Lenses are at x=0, x=10cm, x=20cm. 1. For the first lens (Convex, f₁ = +10 cm), object is at u₁ = -20 cm. 1/+10 = 1/v₁ - 1/(-20) => 1/v₁ = 1/10 - 1/20 = 1/20. So, v₁ =

Q: What are the options for this CUET UG Physics question?

A. 10 cm to the right of third lens · B. 7 cm to the right of third lens · C. 7 cm to the left of third lens · D. at ∞ to right of third lens

Question

Q47. What would be the position of image formed by the lens combination given in the figure below?
f=+10cm
0
20 cm
-10 cm
10 cm
20 cm
+10 cm
(a) 10 cm to the right of third lens
(b) 7 cm to the right of third lens
(c) 7 cm to the left of third lens
(d) at∞ to right of third lens

FirstInTest · Accepted Answer

This problem requires sequential application of the lens formula, 1/f = 1/v - 1/u, where f is focal length, u is object distance, and v is image distance. Distances to the left of the lens are negative, to the right are positive. Lenses are at x=0, x=10cm, x=20cm.
1. For the first lens (Convex, f₁ = +10 cm), object is at u₁ = -20 cm.
   1/+10 = 1/v₁ - 1/(-20) => 1/v₁ = 1/10 - 1/20 = 1/20. So, v₁ = +20 cm. This means the first image (I₁) forms 20 cm to the right of the first lens, i.e., at x = 20 cm.
2. For the second lens (Concave, f₂ = -10 cm), which is at x = 10 cm.
   The image I₁ at x=20 cm acts as the object for the second lens. So, u₂ = (20 cm - 10 cm) = +10 cm (it's a virtual object to the right of lens 2). 
   1/-10 = 1/v₂ - 1/(+10) => 1/v₂ = 1/-10 + 1/10 = 0. So, v₂ = ∞. This means the second image (I₂) forms at infinity.
3. For the third lens (Convex, f₃ = +10 cm), which is at x = 20 cm.
   The image I₂ at infinity acts as the object for the third lens. So, u₃ = -∞ (object from left, at infinity).
   1/+10 = 1/v₃ - 1/(-∞) => 1/v₃ = 1/10. So, v₃ = +10 cm. This means the final image forms 10 cm to the right of the third lens.
Therefore, the final image is formed 10 cm to the right of the third lens. Options (b), (c), and (d) are incorrect.

Answer

10 cm to the right of third lens

Answer

7 cm to the right of third lens

Answer

7 cm to the left of third lens

Answer

at ∞ to right of third lens

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