Question

a convex mirror has a focal length of 40cm. u = 60cm obj height = 10cm h'=? m=? v=?
pls help me I'm in trouble..​

Answer 1

Answer:

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Answer 2

Given :

In convex mirror,

Focal length = 40 cm

Object distance = - 60 cm

Object height = 10 cm

To find :

The image distance, magnification and image height

Solution :

using mirror formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

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

where,

• v denotes Image distance
• u denotes object distance
• f denotes focal length

By substituting all the given values in the formula,

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 60} = \dfrac{1}{40}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{ 60} = \dfrac{1}{40}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{40} + \dfrac{1}{60}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{3 + 2}{120}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{5}{120}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{24}$

$\dashrightarrow\sf v = 24 \: cm$

thus, the image distance is 24 cm

we know that,

The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign that is,

$\dashrightarrow\sf m = - \dfrac{v}{u}$

$\dashrightarrow\sf m = - \dfrac{24}{ - 60}$

$\dashrightarrow\sf m = 0.4$

Thus, the magnification is 0.4 times.

» The Magnification produced by a mirror is equal to the ratio of the height of the image to the height of the object .i.e.,

$\boxed{ \bf m = \dfrac{h'}{h} }$

where,

• h' denotes height of the image
• h denotes height of the object

by, substituting all the given values in the formula,

$\dashrightarrow \sf m = \dfrac{h'}{h}$

$\dashrightarrow \sf 0.4 = \dfrac{h'}{10}$

$\dashrightarrow \sf h' = 0.4 \times 10$

$\dashrightarrow \sf h' = 4 \: cm$

Thus, the image height is 4 cm