JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 17)
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 :
_20th_July_Morning_Shift_en_17_1.png)
55.44 cm
18.23 cm
9.52 cm
37.58 cm
Explanation
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.
$${{{\mu _{II}}} \over v} - {{{\mu _I}} \over u} = {{{\mu _{II}} - {\mu _I}} \over R}$$
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.
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