A planoconcave lens is placed on a paper on which a flower is drawn. How far above its actual position does the flower appear to be?

Question

A planoconcave lens is placed on a paper on which a flower is drawn. How far above its actual position does the flower appear to be? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/eed49369db71badf93f5ef3d32ea054e-.png" /> (A) 10 cm (B) 15 cm (C) 50 cm (D) None of these

Tardigrade Admin · Accepted Answer

Considering refraction at the curved surface,

u = - 20, μ_{2} = 1

μ_{1} = 3/2, R = + 20

Applying

\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}

\Rightarrow \frac{1}{v} - \frac{3/2}{- 20} = \frac{1 - 3/2}{20}

\Rightarrow v = - 10

i.e.,

10 c m

below the curved surface or

10 c m

above the actual position of flower.

Q. A planoconcave lens is placed on a paper on which a flower is drawn. How far above its actual position does the flower appear to be?

10 cm

15 cm

50 cm

None of these

Considering refraction at the curved surface, u=−20,μ2​=1 μ1​=3/2,R=+20 Applying vμ2​​−uμ1​​=Rμ2​−μ1​​ ⇒v1​−−203/2​=201−3/2​ ⇒v=−10 i.e., 10cm below the curved surface or 10cm above the actual position of flower.