Question

find the ratio of height of image formed by concave and convex mirror of focal length 10 cm If object placed at 30 cm

Answer 1

Answer:

Explanation:

for concave mirror , focal length is negative = -10

1/v + 1/u = 1/f

object distance = -30

1/v = 1/f - 1/u

v = -15

magnification -v/u = h {img 1} / h {obj}

h {img 1} / h {obj} = 15/30-----------------eq 1

for convex mirror focal length positive = 10

1/v + 1/u = 1/f

object distance = -30

1/v = 1/f - 1/u

v = 7.5

magnification -v/u = h {img 2} / h {obj}

h {img 2} / h {obj} = 7.5/30--------------------------eq 2

eq 1 / eq 2

h {img 1} / h{img 2} =  2

Answer 2

GiveN :-

A concave mirror and a convex mirror both with focal length 10 cm.

Object distance is 30 cm in both the mirrors.

TO FinD :-

The ratio of height of image formed by concave and height of image formed by convex mirror.

SolutioN :-

For concave mirror,

• u = -30 cm

• f = -10 cm

Applying mirror formula :-

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

$\sf{\implies \dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-30} }\\\\\sf{\implies \dfrac{1}{v}=\dfrac{-1}{10}+\dfrac{1}{30} }\\\\\sf{\implies \dfrac{1}{v}=\dfrac{-3+1}{30} }\\\\\sf{\implies \dfrac{1}{v}=\dfrac{-2}{30} }\\\\\sf{\implies \dfrac{1}{v}=\dfrac{-1}{15} }\\\\\sf{\implies v=-15 }$

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

$\sf{\dfrac{-v}{u}=\dfrac{h_i}{h_o}=\dfrac{-(15)}{-30}=\dfrac{-1}{2} }\\\\\sf{So,\ \dfrac{h_i}{h_o}=\dfrac{-1}{2} }$

Or, hi = -1 and ho = 2.

$\rule{200}{2}$

For convex mirror,

• u = -30 cm

• f = 10 cm

Applying mirror formula :-

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

$\sf{\implies \dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{-30} }\\\\\sf{\implies \dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{30} }\\\\\sf{\implies \dfrac{1}{v}=\dfrac{3+1}{30} }\\\\\sf{\implies \dfrac{1}{v}=\dfrac{4}{30} }\\\\\sf{\implies v=\dfrac{30}{4} }\\\\\sf{\implies v=7.5\ cm}$

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

$\sf{\dfrac{-v}{u}=\dfrac{h_i}{h_o}=\dfrac{-(7.5)}{-30}=\dfrac{7.5}{30}=\dfrac{1}{4} }\\\\\\\\\sf{So,\ \dfrac{h_i}{h_o}=\dfrac{1}{4} }$

Or, hi = 1 and ho = 4.

$\rule{200}{2}$

Now,

ratio of height of image formed by concave and height of image formed by convex mirror = $\sf{\dfrac{-1}{1}}$ = -1 : 1.

$\underline{\rule{290}{2}}$

INDEX :-

• hi is the height of the image.

• ho is the height of the object.