COMEDK2024Evening ShiftPhysicsRay OpticsActual
Figure below shows a lens of refractive index, =1.4 . C₁ and C₂ are the centres of curvature of the two faces of the lens of radii of curvature 4 ~cm and 8 ~cm respectively. The lens behaves as a
Options
- Aconverging lens of focal length 12 ~cm
- Bdiverging lens of focal length 20 ~cm
- Cconverging lens of focal length 20 ~cm
- Ddiverging lens of focal length 12 ~cm
Correct answer
C. converging lens of focal length 20 ~cm
Step-by-step solution
The lens shown is a meniscus lens. We use the lens maker's formula: 1 f = ( - 1) ( 1 R₁ - 1 R₂ ) . For the given lens, the first surface (left) is convex towards the object space, so its center of curvature C₁ is to the right. By sign convention, R₁ = +4 cm . The second surface (right) is also convex towards the object space, so its center of curvature C₂ is also to the right. By sign convention, R₂ = +8 cm . Given = 1.4 , we substitute these values into the formula: 1 f = (1.4 - 1) ( 1 4 - 1 8 ) 1 f = 0.4 ( 2 - 1