Question

An object is placed at a distance of 4 cetiments in front of a concave mirror whose focal length is 5 centiments where is the image formed​

Answer 1

Answer:

$\boxed{\mathfrak{Image \ distance \ (v) = 20 \ m}}$

Given:

Object distance (u) = -4 cm

Focal length (f) = -5 cm

Explanation:

Mirror formula:

$\boxed{ \bf{ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}$

v → Image distance

By substituting values in the formula we get:

$\rm \implies \dfrac{1}{v} - \dfrac{1}{4} = - \dfrac{1}{5} \\ \\ \rm \implies \dfrac{1}{v} = \dfrac{1}{4} - \dfrac{1}{5} \\ \\ \rm \implies \dfrac{1}{v} = \dfrac{5 - 4}{20} \\ \\ \rm \implies \dfrac{1}{v} = \dfrac{1}{20} \\ \\ \rm \implies v = 20 \: m$

Answer 2

Question:-

An object is placed at a distance of 4 cetiments in front of a concave mirror whose focal length is 5 centiments where is the image formed.

Given:-

=> u = -4 ; f = -5 ; image distance = 20m

Formula:-

★ ¹/v + ¹/u = ¹/f

Solution:-

substitute in formula.

➭ 1/v + 1/4 = 1/5

➭ 1 = 5 - 4

v 20

➭ 1/v = 1/20

➭ v = 20

☞ Hence verified

Explanation:-

• The relationship between the radius of curvature and focal length is that the radius of curvature is twice the focal length.In this case, focal length of the concave mirror is given as 5 cm. Hence, the radius of curvature is 5×2=10 cm.

• When the object is placed on the centre of curvature, on the principal axis, in front of a concave lens, the image is formed at the same point. The image formed is real and inverted.

• So, in this case, the object distance is equal to the image distance.Therefore, the object size is equal to the image size.As the object is of 4 cm in length, the image will also be of 4 cm in length.