KCET · Physics · Ray Optics
A luminous point object \(O\) is placed at a distance \(2 R\) from the spherical boundary separating two transparent media of refractive indices \(n_1\) and \(n_2\) as shown, where \(R\) is the radius of curvature of the spherical surface. If \(n_1=\frac{4}{3}, n_2=\frac{3}{2}\) and \(R=10 \mathrm{~cm}\), the image is obtained at a distance from \(P\) equal to
- A 30 cm in the rarer medium
- B 30 cm in the denser medium
- C 18 cm in the rarer medium
- D 18 cm in the denser medium
Answer & Solution
Correct Answer
(A) 30 cm in the rarer medium
Step-by-step Solution
Detailed explanation
From the given diagram in question,
\(n_1=\frac{4}{3} n_2=\frac{3}{2}\)
\(u=-2 R\)
\(\therefore\) Using refraction formula through convex spherical surface.
\(\frac{n_2}{v}-\frac{n_1}{\mu}=\frac{n_2-n_1}{R}\)
\(\Rightarrow \quad \frac{\frac{3}{2}}{v}-\frac{4}{3(-2 R)}=\frac{\frac{3}{2}-\frac{4}{3}}{R} \Rightarrow v=-3 R\)
\(=-3 \times 10 \quad[\because R=10 \mathrm{~cm}]\)
\(=-30 \mathrm{~cm}\)
Since, \(v\) is negative, thus image is obtained at a distance from \(P\) equal to 30 cm in the rarer medium.
\(n_1=\frac{4}{3} n_2=\frac{3}{2}\)
\(u=-2 R\)
\(\therefore\) Using refraction formula through convex spherical surface.
\(\frac{n_2}{v}-\frac{n_1}{\mu}=\frac{n_2-n_1}{R}\)
\(\Rightarrow \quad \frac{\frac{3}{2}}{v}-\frac{4}{3(-2 R)}=\frac{\frac{3}{2}-\frac{4}{3}}{R} \Rightarrow v=-3 R\)
\(=-3 \times 10 \quad[\because R=10 \mathrm{~cm}]\)
\(=-30 \mathrm{~cm}\)
Since, \(v\) is negative, thus image is obtained at a distance from \(P\) equal to 30 cm in the rarer medium.
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