A thin lens of material having refractive index µ = 1.5 has focal length of 20 cm when placed in air. Now it is placed in two different media having refractive indicies µ1 = 1.2 and µ2 = 2.5 convering upper and lower half of the lens as shown in the figure. If the object is at , the sepration between the images is 13x. Write x in your OMR sheet.

Engineering

Physics

Curved Surface Refraction

Question

A thin lens of material having refractive index µ = 1.5 has focal length of 20 cm when placed in air. Now it is placed in two different media having refractive indicies µ₁ = 1.2 and µ₂ = 2.5 convering upper and lower half of the lens as shown in the figure. If the object is at $\infty$ , the sepration between the images is 13x. Write x in your OMR sheet.

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Focal length of upper half is

$F_{1} = (\frac{µ - 1}{µ / µ_{1} - 1}) F_{air} = (\frac{1 .5 - 1}{\frac{1 .5}{1 .2} - 1}) 20 = 40 cm$

Focal length of lower half :

$F_{2} = [\frac{1 .5 - 1}{\frac{1 .5}{2 .5} - 1}] = - 25 cm$

As the object is at $\infty$ . The two images will be at focuses.

Þ separation = |F₁| + |F₂| = 65 cm