Question

An object is placed on the principal axis of a concave mirror of focal length 10 cm at a distance of 15 cm in front of it. Find the nature and the size of the image.​

Answer 1

⟹ v=−30 cm

⟹ v=−30 cmSince v is negative, so image formed is real.

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m=

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v ∴ m=−

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v ∴ m=− −15

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v ∴ m=− −15(−30)

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v ∴ m=− −15(−30)

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v ∴ m=− −15(−30) =−2

⟹ v=−30 cmSince v is negative, so image formed is real.Magnification m= u−v ∴ m=− −15(−30) =−2Since m is greater than 1, so magnified image is formed.

Answer 2

Given:

An object is placed on the principal axis of a concave mirror of focal length 10 cm at a distance of 15 cm in front of it.

To find:

The nature and the size of the image.

Solution:

Focal length of a concave mirror, f = - 10 cm

Distance of a object, u = - 15 cm

Using mirror formula,

♦ 1/f = 1/u + 1/v

Putting the values,

1/-10 = 1/-15 + 1/v

1/v = - 1/10 + 1/15

1/v = (-3 + 2)/30

1/v = - 1/30

v = - 30

Here, v is negative, so the image formed is real image.

M = - v/u

M is the magnificent of a object.

Putting the values,

M = - (-30)/(-15)

M = - (-6)/(-3)

M = - 2

Since M is less than 1, image is smaller than the object. Thus, image formed is diminished.