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

$\underline{\boxed{\it\ \dfrac{1}{f}= [tex]\dfrac{1}{v}+\dfrac{1}{u}}}$ f1=v1+u1

Now , by putting these values in the formula

$\begin{gathered}$$\longrightarrow\sf\ \dfrac{1}{-30}= \dfrac{1}{-50}+\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-1}{30}+\dfrac{1}{50}$

=$\dfrac{1}{v}\\ \\ \\ [tex]\longrightarrow\sf\ \dfrac{-5+3}{150}$

=$\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-2}{150}$

=$\dfrac{1}{v}\\ [tex]\longrightarrow\sf\ v=\cancel{\dfrac{-150}{2}}$

$\longrightarrow\pink{\sf v= (-75)cm}$ \end{gathered}[/tex]

⟶ −301=−501+v1

⟶ 30−1+501=v1

⟶ 150−5+3=v1

⟶ 150−2=v1

⟶ v=2−150

⟶v=(−75)cm

Now we have to find the size of image

By using magnification formula

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

$\begin{gathered}$$[tex]\longrightarrow\sf\ \dfrac{h_i}{15}=[tex]\dfrac{\not{-}(-75)}{\not{-}50}\\ \\ \\ \longrightarrow\sf\ \dfrac{h_i}{15}=[tex]\dfrac{-(75)}{50}\\ \\ \\ [tex]\longrightarrow\sf\ \ h_i= \dfrac{-(\cancel{75}\times 15)}{\cancel{50}}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-3\times 15}{2}\\ \\ \\ [tex]\longrightarrow\sf\ \ h_i= \cancel{\dfrac{-45}{2}}$\\ \\ \\ $\longrightarrow\sf\ \ h_i= (-22.5)\end{gathered}$

⟶ 15hi=−50−(−75)

⟶ 15hi=50−(75)

⟶  hi=50−(75×15)

⟶  hi=2−3×15

⟶  hi=2−45

⟶  hi=(−22.5)

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

Answer 2

Answer:

object distance=p=50cm

Radius of curvature=R=60cm

then f=R/2

f=60/2

f=30cm

image distance=q=?

using mirror formula

1/f=1/p+1/q

1/30=1/50+1/q

1/30–1/50=1/q

(5–3)/150=1/q

2/150=1/q

2q=150

q=75

image distance=75cm