An object of length 2.5 cm is placed at a distance of 1.5 f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principal axis. The length of the image is

5 cm, erect

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 cm, erect

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15 cm, erect

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 cm, inverted

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 5 cm, inverted
By using $\frac{I}{O} = \frac{f}{f - u}$ ; where I =?, $O = + 2.5 c m . f \to - f, u = - 1.5 f$
$∴ \frac{I}{+ 2.5} = \frac{- f}{- f - (- 1.5 f)} \Rightarrow I = - 5 c m$ .(Negative sign indicates that image is inverted.)

Suggest Corrections

Similar questions

Q. An object of length

2 c m

is placed at distance

1.5 f

from a concave mirror where

^{'} f^{'}

is its focal length. Find the length of the image.

An object of height $h_{1} = 2 c m$ is placed perpendicular to the principal axis of a concave mirror of focal length 10cm. The distance of the object from the mirror is 15cm. Find the nature, position and size of the image respectively.

An object is placed at a distance of 10 cm from a convex mirror of focal length 5cm.
Calculate the distance of an image from mirror.

Q. An object of

5 c m

high is placed at a distance

60 c m

in front of a concave mirror of focal length

10 c m

. The size of the image is:

Q. An object of length

4 c m

is placed in front of a concave mirror at a distance of

30 c m

. The focal length of mirror is

15 c m

. Where will the image form?

Join BYJU'S Learning Program

An object of length 2.5 cm is placed at a distance of 1.5 f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principal axis. The length of the image is

The correct option is D 5 cm, invertedBy using IO=ff−u; where I =?, O=+2.5cm.f→−f,u=−1.5f ∴I+2.5=−f−f−(−1.5f)⇒I=−5cm.(Negative sign indicates that image is inverted.)

The correct option is D 5 cm, inverted
By using $\frac{I}{O} = \frac{f}{f - u}$ ; where I =?, $O = + 2.5 c m . f \to - f, u = - 1.5 f$
$∴ \frac{I}{+ 2.5} = \frac{- f}{- f - (- 1.5 f)} \Rightarrow I = - 5 c m$ .(Negative sign indicates that image is inverted.)