Question

An object 2.5 cm high is placed 6.0 cm from a converging lens with a focal

length of 4.0cm. Determine (a) the image distance, (b) the image size and (c)

the magnification.​

Answer 1

Explanation:

(a) The image distance is 12cm

(b) The image size is 5cm

(c) The magnification produced is -2

Given :

An object of 2.5cm height is 6.0cm from a converging lens

A focal length of the lens is 4.0cm

To Find :

The image distance

The image size

The magnification

Formulae used :

Lens formula

\sf \star \: \: \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}⋆

v

1

−

u

1

=

f

1

Formula for magnification of the lens

\sf \star \: \: m = \dfrac{h_{2}}{h_{1}} =\dfrac{v}{u}⋆m=

h

1

h

2

=

u

v

Solution :

Given ,

Object distance , u = -6cm

Height of the object , h1 = 2.5cm

Focal length , f = 4cm

Applying mirror formula ,

\begin{gathered}\sf \implies \dfrac{1}{v}-\dfrac{1}{-6}=\dfrac{1}{4}\\\\ \sf \implies \dfrac{1}{v}+\dfrac{1}{6}=\dfrac{1}{4}\\\\ \sf \implies \dfrac{1}{v}=\dfrac{1}{4} - \dfrac{1}{6}\\\\ \sf \implies \dfrac{1}{v} =\dfrac{1}{12}\\\\ \sf \implies v = 12cm\end{gathered}

⟹

v

1

−

−6

1

=

4

1

⟹

v

1

+

6

1

=

4

1

⟹

v

1

=

4

1

−

6

1

⟹

v

1

=

12

1

⟹v=12cm

Therefore , the image distance is 12cm

Now magnification ,

\begin{gathered}\sf \implies m = \dfrac{v}{u} \\\\ \sf \implies m = \dfrac{12}{-6}\\\\ \sf \implies m = -2\end{gathered}

⟹m=

u

v

⟹m=

−6

12

⟹m=−2

Thus ,the magnification is -2

The negative sign represents image formed is real and inverted

Again we have ,

\begin{gathered}\sf \implies m = \dfrac{h_{2}}{h_{1}}\\\\ \sf \implies -2 = \dfrac{h_{2}}{2.5cm } \\\\ \sf \implies h_{2} = -5cm\end{gathered}

⟹m=

h

1

h

2

⟹−2=

2.5cm

h

2

⟹h

2

=−5cm

The size of the image is 5cm