Question

A 4 CM tall light bulb is placed a distance of 45.7 CM from a concave mirror having focal length of 15.2 CM determine the image distance and size

Answer 1

Given :-

• Height of the light bulb = 4 cm
• Distance of the light bulb from the concave mirror = - 45.7 cm

(object distance is always negative as distances are measured from the pole of the mirror )

• Focal length of the concave mirror = -15.2 cm

( focal length of a concave mirror is always negative )

To find :-

• Image distance
• Size of the image

Solution :-

As per the mirror formula ,

$\bigstar\boxed{\mathtt{\pink{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}$

Now , let's substitute the given values in the above equation ,

$\implies\mathtt{\dfrac{ - 1}{15.2} = \dfrac{1}{v} + \dfrac{ - 1}{45.7}}$

$\implies\mathtt{\dfrac{ - 1}{15.2} = \dfrac{1}{v} - \dfrac{ 1}{45.7}}$

$\implies\mathtt{\dfrac{1}{v} = \dfrac{ - 1}{15.2}+ \dfrac{ 1}{45.7} }$

$\implies\mathtt{\dfrac{1}{v} = \dfrac{ - 45.7 + 15.2}{694.64}}$

$\implies\mathtt{\dfrac{1}{v} = \dfrac{ - 30.5}{694.64}}$

$\implies\mathtt{v = \dfrac{ - 694.64}{ 30.5}}$

$\implies\mathtt{ \red{v = - 22.77 \: cm}}$

The image is placed at a distance of 22.77 cm on the same side as that of the object

_______________

Magnification is given by the formula ,

$\bigstar\boxed{\mathtt{\pink{m = - \dfrac{v}{u} = \dfrac{hi}{ho}}}}$

Now , let's substitute the given values in the above equation ,

$\implies\mathtt{ hi = - \dfrac{ - 22.8}{ - 45.7} \times 4 }$

$\implies\mathtt{ hi = -0.49 \times 4 }$

$\implies\mathtt{ \red{hi = -1.96 \: cm}}$

The height of the image is 1.96 cm .Here the negative sign denotes that the image is inverted and real