Question

Images of same size are formed by a convex lens when an object is placed either at 20 cm or 10 cm distance from the lens. The focal length of the lens is

• A. 12 cm
• B. 40 cm
• C. 18 cm
• D. 15 cm

Hint:

Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \). Magnification: \( m = \frac{v}{u} \). Also \( m = \frac{f}{f+u} \) or \( m = \frac{f-v}{f} \). "Images of same size" means same magnitude of magnification: \( |m_1| = |m_2| \). So, \( \left|\frac{f}{f+u_1}\right| = \left|\frac{f}{f+u_2}\right| \). Given \(u_1, u_2\) are object distances (negative for real objects). This simplifies to \( |f+u_1| = |f+u_2| \). Solve for \(f\).

Answer 1

The Correct Option is D

For a convex lens, magnification \( m = \frac{v}{u} = \frac{f}{f+u} \).
"Images of same size" means the magnitude of magnification \( |m| \) is the same in both cases.
Let \( |m_1| = |m_2| \).
Case 1: Object distance \( u_1 = -20 \) cm.
Case 2: Object distance \( u_2 = -10 \) cm.
Magnification magnitude: \( |m| = \left|\frac{f}{f+u}\right| \).
So, \( \left|\frac{f}{f-20}\right| = \left|\frac{f}{f-10}\right| \).
This implies \( |f-20| = |f-10| \) (assuming \(f \ne 0\)).
This means \( f \) is equidistant from 10 and 20, so \( f = \frac{10+20}{2} = 15 \).
Let's check this logic carefully.
If \( |A|=|B| \), then \(A=B\) or \(A=-B\).
Possibility (a): \( f-20 = f-10 \implies -20 = -10 \), which is false.
Possibility (b): \( f-20 = -(f-10) \) \( f-20 = -f+10 \) \( 2f = 30 \) \( f = 15 \) cm.
Since it's a convex lens, \(f\) should be positive.
\(f=15\) cm is positive.
Let's check the magnifications.
If \( f=15 \) cm: Case 1: \( u_1 = -20 \) cm.
\( m_1 = \frac{15}{15+(-20)} = \frac{15}{-5} = -3 \).
Image is real, inverted, magnified.
\( |m_1|=3 \).
Case 2: \( u_2 = -10 \) cm.
\( m_2 = \frac{15}{15+(-10)} = \frac{15}{5} = 3 \).
Image is virtual, erect, magnified.
\( |m_2|=3 \).
Since \( |m_1|=|m_2| \), the condition "images of same size" is satisfied.
The focal length is 15 cm.
This matches option (4).
Alternative reasoning: For a convex lens, if an object is placed at distance \(u\) and its image is formed at \(v\), then magnification \(m = v/u\).
If \(|m|=k\), then \(|v|=k|u|\).
If the image size is the same as object size, \(|m|=1\), which happens when \(u=2f\).
This question says "images of same size" (plural), meaning the magnifications are same in magnitude for two different object positions.
The condition for same magnitude of magnification \(|m|\) for two object positions \(u_1\) and \(u_2\) for a convex lens occurs when one image is real and the other is virtual, and \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
The property that \(f\) is the arithmetic mean of \(|u_1|\) and \(|u_2|\) when \(|m_1|=|m_2|\) and one image is real and other is virtual.
(This is for \(|m|=1\)).
Not general.
The equation \(|f-u_1|=|f-u_2|\) for \(|f/(f+u_1)| = |f/(f+u_2)|\) implies \(|f+u_1|=|f+u_2|\).
With \(u_1 = -20, u_2 = -10\), we used \(|f-20|=|f-10|\).
This correctly led to \(f=15\).