Question

A mirror is used to produce an image with magnification of $ \frac{1}{4} $. If the distance between object and its image is 40 cm, then the focal length of the mirror is ____.

• A. 10 cm
• B. 12.7 cm
• C. 10.7 cm
• D. 15 cm

Hint:

Remember the sign conventions for object distance (\(u\)), image distance (\(v\)), focal length (\(f\)), and magnification (\(m\)). For real objects, \(u\) is negative. For real images, \(v\) is negative; for virtual images, \(v\) is positive. For concave mirrors, \(f\) is negative; for convex mirrors, \(f\) is positive. Magnification \(m = -\frac{v}{u}\); positive \(m\) indicates a virtual and erect image, while negative \(m\) indicates a real and inverted image.

Answer 1

The Correct Option is C

Step 1: Understand the given information.
Magnification, \( m = \frac{1}{4} \) (positive, so the image is virtual and erect, implying a convex mirror, or real and inverted) 
Distance between object and image, \( |v - u| = 40 \) cm 
Step 2: Consider the case of a real and inverted image (concave mirror).
For a real and inverted image, \( m = -\frac{v}{u} = -\frac{1}{4} \), so \( v = \frac{1}{4} u \). 
Since the image is real and inverted, it forms on the same side as the incident light, so \( v \) is negative according to the sign convention if \( u \) is negative. 
Let \( u = -x \) where \( x>0 \). Then \( v = -\frac{1}{4} x \). 
The distance between object and image is \( |v - u| = |-\frac{1}{4} x - (-x)| = |-\frac{1}{4} x + x| = |\frac{3}{4} x| = 40 \). 
So, \( \frac{3}{4} x = 40 \implies x = \frac{160}{3} \) cm. 
Therefore, \( u = -\frac{160}{3} \) cm and \( v = -\frac{1}{4} \times \frac{160}{3} = -\frac{40}{3} \) cm. 
Using the mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-40/3} + \frac{1}{-160/3} = -\frac{3}{40} - \frac{3}{160} = \frac{-12 - 3}{160} = -\frac{15}{160} = -\frac{3}{32} \). So, \( f = -\frac{32}{3} \approx -10.67 \) cm. 
The focal length is positive for a concave mirror, so there is a sign error somewhere. 
Let's recheck the magnification sign. If \( m = -\frac{v}{u} = -\frac{1}{4} \), then \( v = \frac{1}{4} u \). 
For a real image, \( v \) and \( u \) have the same sign. With the convention that real objects have negative \( u \), real images have negative \( v \). So, \( v = \frac{1}{4} u \implies v \) is less negative than \( u \), which is consistent with a real image formed by a concave mirror. 
Distance \( |v - u| = |\frac{1}{4} u - u| = |-\frac{3}{4} u| = 40 \). Since \( u \) is negative, \( -\frac{3}{4} u = 40 \implies u = -\frac{160}{3} \) cm. 
Then \( v = \frac{1}{4} (-\frac{160}{3}) = -\frac{40}{3} \) cm. \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-40/3} + \frac{1}{-160/3} = -\frac{3}{40} - \frac{3}{160} = \frac{-12 - 3}{160} = -\frac{15}{160} = -\frac{3}{32} \). \( f = -\frac{32}{3} \approx -10.67 \) cm. 
The focal length of a concave mirror is negative. The magnitude is \( 10.67 \) cm. 
Step 3: Consider the case of a virtual and erect image (convex mirror).
For a virtual and erect image, \( m = -\frac{v}{u} = \frac{1}{4} \), so \( v = -\frac{1}{4} u \). 
For a real object, \( u \) is negative. Let \( u = -x \) where \( x>0 \). Then \( v = \frac{1}{4} x \). 
The image is virtual, so \( v \) is positive. Distance \( |v - u| = |\frac{1}{4} x - (-x)| = |\frac{5}{4} x| = 40 \). 
So, \( \frac{5}{4} x = 40 \implies x = 32 \) cm. 
Therefore, \( u = -32 \) cm and \( v = \frac{1}{4} (32) = 8 \) cm. 
Using the mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{8} + \frac{1}{-32} = \frac{4 - 1}{32} = \frac{3}{32} \). 
So, \( f = \frac{32}{3} \approx 10.67 \) cm. 
The focal length of a convex mirror is positive. 
The magnitude of the focal length is approximately 10.7 cm.