Question

The point C denotes the centre of curvature of the curved surface of refractive index 1.5. An object O is placed in air at a distance of 15 cm. Location of the image will be –

• A. – 30 cm
• B.
  • 30 cm
• C. – 20 cm
• D.
  • 20 cm

Answer 1

Answer: (A)

– 30 cm

Answer 2

Here u = – 15 cm, R= + 30 cm, m₁ = 1 and m₂ = 1.5

= Ž – $\frac { 1 } { - 15 }$ =

Ž + $\frac { 1 } { 15 }$= $\frac { 0.5 } { 30 }$ = $\frac { 1 } { 60 }$

Ž $\frac { 1.5 } { \mathrm { v } }$ = $\frac { 1 } { 60 }$ – $\frac { 1 } { 15 }$

Ž v = – 20 × 1.5 = – 30 cm

so image is formed 30 cm left to the spherical surface and is virtual erect.