Question

An object of height 7.5 cm is placed in front of
a convex mirror of radius of curvature 25 cm a
a distance of 40 cm. The height of the image
should be​

Answer 1

Answer:

size of object, h = 7.5cm

radius of curvature of convex mirror, R = +25cm

focal length, f = R/2 = 25/2 cm

object distance, u = -40cm

image distance = v

we know that

\begin{lgathered}\frac{1}{v} + \frac{1}{u}= \frac{1}{f} \\ \\ \Rightarrow \frac{1}{v}= \frac{1}{f} - \frac{1}{u}\\ \\ \Rightarrow \frac{1}{v}= \frac{1}{25/2} - \frac{1}{40}\\ \\ \Rightarrow \frac{1}{v}= \frac{2}{25} - \frac{1}{40}\\ \\ \Rightarrow \frac{1}{v}= \frac{16-5}{200}\\ \\ \Rightarrow \frac{1}{v}= \frac{11}{200}\\ \\ \Rightarrow v= \frac{200}{11}cm\end{lgathered}

v

1

+

u

1

=

f

1

⇒

v

1

=

f

1

−

u

1

⇒

v

1

=

25/2

1

−

40

1

⇒

v

1

=

25

2

−

40

1

⇒

v

1

=

200

16−5

⇒

v

1

=

200

11

⇒v=

11

200

cm

magnification = \frac{h'}{h} =- \frac{v}{u}

h

h

′

=−

u

v

h' = height of image

\begin{lgathered}\frac{h'}{h} =- \frac{v}{u}\\ \\ h' = -h \times \frac{v}{u}=-7.5 \times \frac{200}{11 \times (-40)}= \frac{75}{22}\ cm=3.41cm\end{lgathered}

h

h

′

=−

u

v

h

′

=−h×

u

v

=−7.5×

11×(−40)

200

=

22

75

cm=3.41cm

Size of image is 3.41cm.