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
[amp_mcq option1=”real and magnified” option2=”virtual and magnified” option3=”real and reduced” option4=”virtual and reduced” correct=”option1″]
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.