Question 131: Ray Optics

Question

Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :

Accepted Answer

As we know that, the equation of refraction at spherical surface is :
$${{{\mu _{II}}} \over v} - {{{\mu _I}} \over u} = {{{\mu _{II}} - {\mu _I}} \over R}$$
where,
$$\mu$$
II
= refractive index of region II = 1.4
$$\mu$$
I
= refractive index of region I = 1.25
R = radius of curvature = $$-$$25 cm
u = object distance = $$-$$40 cm
and v = image distance.
$$ \Rightarrow {{1.4} \over v} - {{1.25} \over { - 40}} = {{1.4 - 1.25} \over { - 25}}$$
$$ \Rightarrow {{1.4} \over v} = {{ - 0.15} \over {25}} - {{1.25} \over {40}}$$ $$\Rightarrow$$ v = $$-$$ 37.58 cm
where, negative sign indicate real image.

Answer

55.44 cm

Answer

18.23 cm

Answer

9.52 cm

Answer

37.58 cm

Ray Optics

Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :

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