Question

A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12 cm with respect to air. An object is situated in the denser medium at a distance of 9 cm from the pole of the surface. Locate the image due to refraction in air​.
(A) A real image at 8 cm
(B) A virtual image at 8 cm
(C) A real image at 4.8 cm
(D) A virtual image at 4.8 cm

Answer 1

Hint: For finding the location of image, we would be using the formula \[\dfrac{{{\mu }_{1}}}{v}-\dfrac{{{\mu }_{2}}}{u}=\dfrac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\] for refraction at the spherical surface. A Refractive index or you can call it an index of refraction is defined as the measure of the bending of a ray of light. While passing from one medium into another medium.

Formula Used: For refraction at the spherical surface. We are using the formula :-
\[\dfrac{{{\mu }_{1}}}{v}-\dfrac{{{\mu }_{2}}}{u}=\dfrac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\]
Where\[{{\mu }_{1}}\]is the refractive index of that medium into which the light rays are entering or you can say it as a rarer medium.
\[{{\mu }_{2}}\]​ is the refractive index of that medium from which the light rays are coming or you can say it as a denser medium.
Also, R is the radius of curvature of the spherical surface.
While, \[\mu \]and \[v\]is the object distance and image distance respectively.

Complete step-by-step solution:
It is given that,
\[u=-9~cm\]
\[{{\mu }_{1}}=1\](air)
\[{{\mu }_{2}}=1.5\]
\[R=-12~cm\]
After, putting all the value. We get,
\[\Rightarrow \dfrac{1}{v}-\dfrac{1.5}{-9}=\dfrac{1-1.5}{-12}\]
\[\Rightarrow \dfrac{1}{v}=\dfrac{1-1.5}{-12}+\dfrac{1.5}{-9}\]
\[\Rightarrow \dfrac{1}{v}=\dfrac{-0.5}{-12}+\dfrac{-1.5}{9}\]
\[\Rightarrow \dfrac{1}{v}=\dfrac{1}{24}-\dfrac{3}{18}\]
\[\Rightarrow \dfrac{1}{v}=\dfrac{1}{24}-\dfrac{1}{6}\]
\[\Rightarrow \dfrac{1}{v}=\dfrac{1-4}{24}\]
\[\Rightarrow \dfrac{1}{v}=\dfrac{-3}{24}\]
$\Rightarrow v=\dfrac{-24}{3}$
\[\therefore v=-8\]
Finally, \[v=-8\]cm
Thus, the result we get is that a virtual image is formed on the same side as the object but it is formed at a distance of 8 cm from the pole.
Hence, the correct answer for the location of the image formed due to the refraction in the air​ is option B which is A virtual image at 8 cm.

Note: Don't get confused while solving equations. Always prefer solving equation in a stepwise manner as well as always take\[{{\mu }_{1}}\]for that medium into which the light rays are entering that is rarer medium and​ \[{{\mu }_{2}}\]for that medium from which the light rays are coming, that is denser medium to get rid of confusions as well as problems.