A point object is placed on the principal axis of a concave mirror, at a distance of 15 cm from the pole. The radius of curvature of the mirror is 20 cm and the object is made to oscillate along the principal axis with an amplitude of 2 mm . The amplitude of its image will be

Question

A point object is placed on the principal axis of a concave mirror, at a distance of 15 cm from the pole. The radius of curvature of the mirror is 20 cm and the object is made to oscillate along the principal axis with an amplitude of 2 mm . The amplitude of its image will be (A) 2mm (B) 4mm (C) 8mm (D) 16mm

Tardigrade Admin · Accepted Answer

Using mirror formula, (1/- 1 0) = (1/ textv) + (1/- 1 5) ⇒ textv = - 3 0 cm |Axial magnification mid=( mathrmV2/ mathrmu2)=((30/15))2=4 Amplitude of image = 4 x 2 = 8 mm.

Q. A point object is placed on the principal axis of a concave mirror, at a distance of $15 c m$ from the pole. The radius of curvature of the mirror is $20 c m$ and the object is made to oscillate along the principal axis with an amplitude of $2 mm$ . The amplitude of its image will be

$2 mm$

$4 mm$

$8 mm$

$16 mm$

Using mirror formula,
$\frac{1}{- 10} = \frac{1}{v} + \frac{1}{- 15} \Rightarrow v = - 30 c m$
|Axial magnification $∣= \frac{V ^{2}}{u ^{2}} = (\frac{30}{15})^{2} = 4$
Amplitude of image = 4 x 2 = 8 mm.

Q. A point object is placed on the principal axis of a concave mirror, at a distance of 15cm from the pole. The radius of curvature of the mirror is 20cm and the object is made to oscillate along the principal axis with an amplitude of 2mm . The amplitude of its image will be

2mm

4mm

8mm

16mm