Question

Light from a small object in air falls on a spherical glass surface $(n = 1.5)$ of radius of curvature $R$. A real image of the object will be formed if the object distance $u$ is related to $R$ as:

• A. $u < \dfrac{R}{2}$
• B. $\dfrac{R}{2} < u < R$
• C. $R < u < 2R$
• D. $u > 2R$

Hint:

Use \(\dfrac{1.5}{v} - \dfrac{1}{u} = \dfrac{0.5}{R}\) with \(u\) negative. A positive \(v\) needs \(|u| > 2R\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Light goes from air (\(n_1 = 1\)) into glass (\(n_2 = 1.5\)) at a spherical surface of radius \(R\). I take the surface as convex towards the object, so \(R\) is positive, and use the Cartesian sign convention.

Step 2: Key Formula:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
The object is on the left, so \(u\) is negative: \(u = -|u|\). A real image has \(v > 0\).

Step 3: Put in the Numbers:
\[ \frac{1.5}{v} + \frac{1}{|u|} = \frac{0.5}{R} \]
\[ \frac{1.5}{v} = \frac{0.5}{R} - \frac{1}{|u|} \]

Step 4: Condition for a Real Image:
For a real image, \(v\) is positive, so the right side must be positive.
\[ \frac{0.5}{R} > \frac{1}{|u|} \quad \Rightarrow \quad |u| > 2R \]
Check: at \(|u| \to \infty\), \(v = 3R\). As the object comes closer, \(v\) grows and becomes infinite at \(|u| = 2R\). Inside \(2R\), \(v\) turns negative (virtual image).

Step 5: Why the Other Options Are Wrong:
For \(R < u < 2R\), \(R/2 < u < R\) and \(u < R/2\), the object lies closer than \(2R\). In all of these the right side is negative, so the image is virtual.

Final Answer:
A real image forms only if the object distance is greater than \(2R\). This is option (D). \[ \boxed{u > 2R} \]