Question

If a man face is 25cm in front of concave shaving mirror producing erect image 1.5 times the size of face, focal length of the mirror would be :
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Answer 1

Answer:

The focal length of the mirror is 75 cm

Explanation:

Given :

A man's face is 25 cm in front of concave shaving mirror producing erect image 1.5 times the size of face.

To find :

the focal length of the mirror

Solution :

object distance, u = -25 cm

size of the image = 1.5 × size of the face

 $\sf h_i=1.5h_o \\\\ \sf \dfrac{h_i}{h_o}=1.5$

Magnification is the ratio of size of the image to that of the object.

$\longmapsto \sf m=\dfrac{-v}{u}=\dfrac{h_i}{h_o}$

 $\implies \sf 1.5=\dfrac{-v}{-25} \\\\ \implies \sf 1.5=\dfrac{v}{25} \\\\ \implies \sf v=1.5 \times 25 \\\\ \implies \sf v=\dfrac{15}{10} \times 25 \\\\ \implies \sf v=\dfrac{75}{2} cm$

image distance, v = 75/2 cm

Now, applying mirror formula,

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

Substitute,

 $\sf \dfrac{1}{f}=\dfrac{1}{\dfrac{75}{2}} +\dfrac{1}{-25} \\\\ \sf \dfrac{1}{f}=\dfrac{2}{75}-\dfrac{1}{25} \\\\ \sf \dfrac{1}{f}=\dfrac{2}{75}-\dfrac{3}{75} \\\\ \sf \dfrac{1}{f}=\dfrac{2-3}{75} \\\\ \sf \dfrac{1}{f}=\dfrac{-1}{75} \\\\ \sf f=-75 \ cm$

Therefore, the focal length of the mirror is 75 cm.