Question

Mirror: Concave mirror; u = - 25 cm; v = - 15 cm; then magnification of the image is​

Answer 1

Explanation:

An object is placed at a distance of 25 cm from a concave mirror focal length of 15 cm. What is the distance for the image from the mirror?

We have our mirror formula:

1/f = 1/v + 1/u, where f is focal length, v is distance from pole to image and u is distance from pole to object.

By sign convention, when distance from pole to image/object is measured, it is positive if it is measured in the same direction as the incident ray of light hitting the mirror and negative if it is measured in the direction opposite to the incident ray. Also by sign convention, the focal length of a concave mirror is negative (that of a convex mirror is positive).

In this case, we have an object placed with a concave mirror 25cm away from it to the right, like this:

∆ ←————25cm————→)

Therefore substituting values:

U=-25cm

v=?

f=-15cm

We get: -1/15 = 1/v - 1/25

=> v = -75/2 cm = -37.5cm

Which means the image is 75/2 cm to the left of the concave mirror.

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Answer 2

Provided that:

• Concave mirror is given.
• Distance of object = -25 cm
• Image distance = -15 cm

To calculate:

• Magnification of image

Solution:

• Magnification of image = -0.6

Using concept:

• Magnification formula

Using formula:

• Magnification formula:

• ${\small{\underline{\boxed{\pmb{\sf{m = \dfrac{h'}{h} = \dfrac{-v}{u}}}}}}}$

Where, m denotes magnification, h′ denotes height of the image, h denotes height of the object, v denotes image distance and u denotes object distance.

Knowledge required:

• If the magnification produced by a spherical mirror is in negative then the mirror is always “Concave Mirror.”

• If the magnification produced by a spherical mirror is in positive then the mirror is always “Convex Mirror.”

• If magnification is negative in a concave mirror then it's nature is “Real and Inverted” always.

• If magnification is positive then it's nature is “Virtual and Erect” always.

• If in the ± magnification, magnitude > 1 then the image formed is “Enlarged”.

• If in the ± magnification, magnitude < 1 then the image formed is “Diminished”.

• If in the ± magnification, magnitude = 1 then the image formed is “Same sized”.

Required solution:

~ By using magnification formula let us calculate the magnification of image.

$:\implies \sf m = \dfrac{h'}{h} = \dfrac{-v}{u} \\ \\ :\implies \sf m = \dfrac{-v}{u} \\ \\ :\implies \sf m = \dfrac{-(-15)}{-25} \\ \\ :\implies \sf m = \dfrac{+15}{-25} \\ \\ :\implies \sf m = \dfrac{+3}{-5} \\ \\ :\implies \sf m = -0.6 \\ \\ :\implies \sf Magnification = -0.6 \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

Therefore,

The image is real and inverted!

The image is diminished!