Two concave glass refracting surfaces, each with radius of curvature R = 35 cm and refractive index μ = 1.5, are placed facing each other in air as shown in Fig. A point object O is placed at a distance of R/2 from one of the surfaces as shown. Find the separation between the images of O formed by each refracting surface.

Solution

Given, two concave glasses.
Radius of curvature of each glass, R = 35 cm
Refractive index = 1.5

For refraction at the first surface, we have
Object distance,

u = AO = - \frac{3 R}{2} = - \frac{105}{2} cm

μ_{2} = 1.5 μ_{1} = 1 Radius of curvature, R = - 35 cm

The distance v₁ of the image I₁ from A is given by the relation

\frac{μ_{2}}{v_{1}} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}

\Rightarrow

\frac{1.5}{v_{1}} - \frac{1}{- 105 / 2} = \frac{1.5 - 1}{- 35}

\Rightarrow v_{1} = - 45 cm i . e ., {AI}_{1} = 45 cm .

Since the radius of curvature is 35 cm, this image is 10 cm to the right of P.
i.e., PI₁ = 10 cm.

For refraction at second surface we have

Object distance,

u = BO = - \frac{R}{2} = - \frac{35}{2} cm

The distance v₂, of the image I₂ , from the second surface is given by

\frac{μ_{2}}{v_{2}} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}

which gives v₂ = – 21 cm or BI₂ = 21 cm.

Thus, the image is 14 cm to the right of P.
i.e., PI₂ = 14 cm.

Hence, the separation between the two images (I, I₂) = 14 – 10 = 4 cm.

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