Question

If an object is placed 50 cm in front of a concave mirror of 60 cm radius, what is the image size if object size is 15cm?

Answer 1

Given :-

• Object distance (u)= -50cm

• Radius (R)= -60cm          [•°• Concave mirror]

• focal length (f)= R/2= -60/2= (-30)cm

• Size of object = 15cm

To Find :-

We have to find the size of image

Solution :-

Using mirror formula

$\underline{\boxed{\it\ \dfrac{1}{f}= \dfrac{1}{v}+\dfrac{1}{u}}}$

Now , by putting these values in the formula

$\longrightarrow\sf\ \dfrac{1}{-30}= \dfrac{1}{-50}+\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-1}{30}+\dfrac{1}{50}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-5+3}{150}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-2}{150}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ v= \cancel{\dfrac{-150}{2}}\\ \\ \\ \longrightarrow\purple{\sf v= (-75)cm}$

Now we have to find the size of image

By using magnification formula

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

$\longrightarrow\sf\ \dfrac{h_i}{15}=\dfrac{\not{-}(-75)}{\not{-}50}\\ \\ \\ \longrightarrow\sf\ \dfrac{h_i}{15}=\dfrac{-(75)}{50}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-(\cancel{75}\times 15)}{\cancel{50}}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-3\times 15}{2}\\ \\ \\ \longrightarrow\sf\ \ h_i= \cancel{\dfrac{-45}{2}}\\ \\ \\ \longrightarrow\sf\ \ h_i= (-22.5)$

$\underline{\bigstar{\textsf{\textbf{\ Size\ of\ image = (-22.5)cm}}}}$

Answer 2

Answer:

Given :-

Object distance = -50cm

Radius = -60cm

Size of object = 15cm

To Find :-

Size of image

Solution :-

At first

F = R/2

F = Focal length

R = Radius

F = -60/2

F = -30 cm

Now,

1/u + 1/v = 1/f

1/-50 + 1/v = 1/-30

1/v = 1/-50 - 1/-30

1/v = -1/50 + 1/30

1/v = -2/150

v = -75 cm

Now,

m = hi/ho = -v/u

-(-75)/-50 = hi/15

-75/50 = hi/15

-15/10 = hi/15

-15 × 15/10

-15 × 3/2

-45/2

-22.5

Size of image = -22.5 cm