Solve this question---**Question Stem:**
An extended object is placed at point O, 10 cm in front of a convex lens L₁ and a concave lens L₂ is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens system is
**Chart Description:**
The figure shows a simple optical system along a horizontal optical axis.
- An object, represented by an upward arrow labeled 'O', is placed on the optical axis to the left.
- A convex lens (L₁) is placed on the optical axis to the right of the object.
- A concave lens (L₂) is placed on the optical axis to the right of the convex lens.
- The distance between the object 'O' and the center of lens L₁ is indicated by a double arrow and labeled "10 cm".
- The distance between the center of lens L₁ and the center of lens L₂ is indicated by a double arrow and labeled "10 cm".
**Options:**
(A) 0.4
(B) 0.8
(C) 1.3
(D) 1.6
**Other Relevant Text:**
L₁
L₂
O
10 cm
10 cm
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We have an optical system with an object O placed 10 centimeters in front of a convex lens L1. A concave lens L2 is positioned 10 centimeters behind L1. Both lenses have radii of curvature of 20 centimeters and refractive index of 1.5. We need to find the total magnification of this lens system.
First, we calculate the focal lengths of both lenses using the lens maker's formula. For the convex lens L1 with R1 equals positive 20 centimeters and R2 equals negative 20 centimeters, we get f1 equals positive 20 centimeters. For the concave lens L2 with R1 equals negative 20 centimeters and R2 equals positive 20 centimeters, we get f2 equals negative 20 centimeters.
Now we find the image formed by the first lens L1. Using the lens formula with object distance u1 equals negative 10 centimeters and focal length f1 equals 20 centimeters, we calculate v1 equals negative 20 centimeters. This means the image is formed 20 centimeters to the left of L1. The magnification by L1 is m1 equals 2, making the image twice as large as the object.
Now the image from L1 becomes the object for L2. Since the image from L1 is 20 centimeters to the left of L1, and L2 is 10 centimeters to the right of L1, the object distance for L2 is u2 equals negative 30 centimeters. Using the lens formula with f2 equals negative 20 centimeters, we get v2 equals negative 12 centimeters. The magnification by L2 is m2 equals 0.4.
Finally, we calculate the total magnification of the lens system. The total magnification is the product of individual magnifications: M total equals m1 times m2 equals 2 times 0.4 equals 0.8. Therefore, the answer is B, 0.8.