Solveeit Logo
Login

Question

Question: A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12 cm with resp...

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

Explanation

Solution

For finding the location of image, we would be using the formula μ1v−μ2u=μ1−μ2R\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 :-
μ1v−μ2u=μ1−μ2R\dfrac{{{\mu }_{1}}}{v}-\dfrac{{{\mu }_{2}}}{u}=\dfrac{{{\mu }_{1}}-{{\mu }_{2}}}{R}
Whereμ1{{\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.
μ2{{\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, μ\mu and vvis the object distance and image distance respectively.

Complete step-by-step solution:
It is given that,
u=−9 cmu=-9~cm
μ1=1{{\mu }_{1}}=1(air)
μ2=1.5{{\mu }_{2}}=1.5
R=−12 cmR=-12~cm
After, putting all the value. We get,
⇒1v−1.5−9=1−1.5−12\Rightarrow \dfrac{1}{v}-\dfrac{1.5}{-9}=\dfrac{1-1.5}{-12}
⇒1v=1−1.5−12+1.5−9\Rightarrow \dfrac{1}{v}=\dfrac{1-1.5}{-12}+\dfrac{1.5}{-9}
⇒1v=−0.5−12+−1.59\Rightarrow \dfrac{1}{v}=\dfrac{-0.5}{-12}+\dfrac{-1.5}{9}
⇒1v=124−318\Rightarrow \dfrac{1}{v}=\dfrac{1}{24}-\dfrac{3}{18}
⇒1v=124−16\Rightarrow \dfrac{1}{v}=\dfrac{1}{24}-\dfrac{1}{6}
⇒1v=1−424\Rightarrow \dfrac{1}{v}=\dfrac{1-4}{24}
⇒1v=−324\Rightarrow \dfrac{1}{v}=\dfrac{-3}{24}
⇒v=−243\Rightarrow v=\dfrac{-24}{3}
∴v=−8\therefore v=-8
Finally, v=−8v=-8cm
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 takeμ1{{\mu }_{1}}for that medium into which the light rays are entering that is rarer medium and​ μ2{{\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.