A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is,

Question

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is, (A) 2.5 cm (B) 5 cm (C) 10 cm (D) 15 cm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-dcfi42pqw0vtkes7.jpg" /> from mirror formula (1/v)+(1/u)=(1/f) image distance of A (1/vA)+(1/(- 30))=(1/- 10)⇒ vA=-15 cm Also image distance of C,vc=-20 cm The length of image =|vA - vc|=|.-15-(.-20.)|.=5 cm

Q. A rod of length $10 c m$ lies along the principal axis of a concave mirror of focal length $10 c m$ in such a way that its end closer to the pole is $20 c m$ away from the mirror. The length of the image is,

$2.5 c m$

$5 c m$

$10 c m$

$15 c m$

from mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ image distance of $A$
$\frac{1}{v _{A}} + \frac{1}{( - 30 )} = \frac{1}{- 10} \Rightarrow v_{A} = - 15 c m$
Also image distance of $C, v_{c} = - 20 c m$
The length of image
$= ∣ v_{A} - v_{c} ∣ = ∣ - 15 - (- 20) ∣ = 5 c m$

Q. A rod of length 10cm lies along the principal axis of a concave mirror of focal length 10cm in such a way that its end closer to the pole is 20cm away from the mirror. The length of the image is,

2.5cm

5cm

10cm

15cm