Question

find the position and nature of the image of an object of height 3 cm when placed 60 cm from a convex mirrorof focal length 15 cm

Answer 1

GIVEN :-

⟿ Height of the object ($h_i$) = 3 cm

⟿ Object distance (u) = -60 cm (Negative as it is placed in front of the mirror)

⟿ Focal length (f) = 15 cm (Positive, as it is a convex mirror)

TO FIND :-

☞ The position and nature of the object.

SOLUTION :-

We know,

$\displaystyle{\sf{\frac{1}{v}+\frac{1}{u}=\frac{1}{f} }}$

$\displaystyle{\leadsto \sf{\frac{1}{v}=\frac{1}{f} -\frac{1}{u} }}$

Substituting the known values :-

$\displaystyle{\sf{\implies \frac{1}{v}=\frac{1}{15}-\frac{1}{(-60)} }}\\\\\displaystyle{\sf{\implies \frac{1}{v}=\frac{1}{15}+\frac{1}{60} }}\\\\\displaystyle{\sf{\implies \frac{1}{v}=\frac{4+1}{60} }}\\\\\displaystyle{\sf{\implies \frac{1}{v}=\frac{5}{60} }}\\\\\displaystyle{\sf{\implies \frac{1}{v}=\frac{1}{12} }}\\\\\underline{\boxed{\displaystyle{\sf{\implies v=12\ cm}}}}$

⥇ So, the position of the image is 12 cm behind the mirror(as the magnitude is positive).

Now, magnification = -v/u

→ M = -(12/-60)

→ M = 1/5

→ M = 0.2

⥇ As the magnification is positive, then the image is virtual and erect.

We also know,

$\displaystyle{\sf{\frac{h_i}{h_o}=M }}\ \textsf{(Height of the image/Height of the object = Magnification)}$

$\displaystyle{\sf{\implies \frac{3}{h_o}=0.2 }}\\\\\boxed{\displaystyle{\sf{\implies h_o=15\ cm}}}$

Or, size of the object = 15 cm.

⥇ As ho > hi, so, the image is diminished.

Answer 2

Answer :

The image will form 12cm behind the mirror

and the image will be highly diminished i.e. its image size is 0.6cm and also it will be virtual and erect

Given :

• The height of the object is 3cm
• The object is places 60cm from the convex mirror
• The focal length of the convex mirror is 15cm

Formula to be Used :

Mirror formula :

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

Magnification of a mirror :

$\star \: \: \bf{m = \dfrac{h _{2} }{h_{1}} } = - \dfrac{v}{u}$

Solution :

Let us consider the following :

• The focal length be f

• The object distance be u

• The image distance be v

• The size object be h₁

• The size of image be h₂

From given data ,

f = 15cm

u = -60cm

h₁ = 3cm

Using the mirror formula :

$\sf \implies \dfrac{1}{v} + \dfrac{1}{ - 60} = \dfrac{1}{15} \\ \\ \implies \sf \dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{60} \\ \\ \implies \sf \dfrac{1}{v} = \frac{4 + 1}{60} \\ \\ \sf \implies \sf \dfrac{1}{v} = \frac{5}{60} \\ \\ \implies \sf \dfrac{1}{v} = \frac{1}{12} \\ \\ \implies \bf v = 12$

Therefore, the image distance is +12cm

i.e. image will form 12cm behind the mirror

Now magnification :

$\sf \implies{m = - \dfrac{12}{ - 60} } = \dfrac{ h_{2}}{ 3 } \\ \\ \implies \sf\dfrac{12}{60} = \frac{ h_{2} }{3} \\ \\ \implies \sf h_{2} = \dfrac{12}{60} \times 3 \\ \\ \sf \implies h_{2} = \frac{3}{5} \\ \\ \bf \implies h_{2} = 0.6$

Thus , the size of image is 0.6 cm

Thus , the image will be virtual , erect and diminished