An object is placed at the centre of curvature of a concave mirror of focal length 16 cm. If the object is shifted by 8 cm towards the focus, the nature of the image would be
real and magnified
virtual and magnified
real and reduced
virtual and reduced
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-1 – 2016
The object is shifted by 8 cm towards the focus (which is at 16 cm). The new object distance is u’ = 32 cm – 8 cm = 24 cm.
The object is now located between the centre of curvature (32 cm) and the focus (16 cm). For a concave mirror, when the object is placed between C and F, the image formed is real, inverted, and magnified.
– Object at C: Image at C, real, inverted, same size.
– Object between C and F: Image beyond C, real, inverted, magnified.
1/16 = 1/v + 1/24
1/v = 1/16 – 1/24 = (3 – 2)/48 = 1/48
v = 48 cm.
Since v is positive, the image is real and formed in front of the mirror, beyond C (at 32 cm).
The magnification m = -v/u = -48/24 = -2.
The negative sign indicates an inverted image, and |m| > 1 indicates a magnified image. Thus, the image is real, inverted, and magnified.