Question

An object 5cm in length is placed at a distance of 20 cm in front of a concave mirror of radius of curvature 30 cm. Find the nature , position and size of object.

Answer 1

Given

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• height of object, h₁ = 5 cm
• position of object, u = -20 cm
• Mirror given is concave mirror
• radius of curvature, r = -30 cm

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To find

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• nature of image
• position of image, v
• size of image , h₂

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Formulae used

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➤ Relation b/w radius of curvature and focal length of mirror

Radius of curvature is equal to 2 times focal length of mirror

$\red{ \bigstar} \boxed{ \sf{r = 2f}}$

( where r is the radius of curvature and f is the focal length of mirror)

( Focal length of concave mirror is always Negative )

➤ Mirror formula

$\red{ \bigstar}\boxed{ \sf{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }}$

( where f is the focal length, v is the position of image, u is the position of object)

➤ formula for Magnification of mirror

$\red{ \bigstar}\boxed{ \sf{m = \frac{ - v}{u} = \frac{h_2}{h_1} }}$

(where m is the magnification, vi s the position of image , u is the position of object, h₁ is the height of object , h₂ is the height of image )

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Solution

Finding the focal length of concave lens

$\implies \sf{r = 2f} \\ \implies \sf{ - 30 = 2f} \\ \underline{ \underline{ \implies \red{ \sf{f = - 15 \: cm}}}}$

Finding the position of image

Using mirror formula

$\implies \sf{ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} } \\ \\ \implies \sf{ \frac{1}{ - 15} = \frac{1}{ - 20} + \frac{1}{v} } \\ \\ \implies \sf{ \frac{1}{v} = \frac{1}{ - 15} + \frac{1}{20} } \\ \\ \implies \sf{ \frac{1}{v} = \frac{ - 4 + 3}{60} } \\ \\ \underline{ \underline{ \implies \red{ \sf{v = - 60 \: cm}}}}$

Finding the Height of image

Using formula for magnification

$\implies \sf{ \frac{ - v}{u} = \frac{h_2}{h_1} } \\ \\ \implies \sf{ \frac{ - ( - 60)}{ - 20} = \frac{h_2}{5} } \\ \\ \underline{ \underline{\implies \red{ \sf{h = - 15 \: cm}}}}$

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Hence,

✤ Nature of image is Real and inverted.

( because image is formed in front of mirror (v= -60) and height of image is found negative)

ᅠ

✤ Position of image is in front of mirror at v = - 60 cm

ᅠ

✤ Size of image is 15 cm ( and inverted image).

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Answer 2

Question:-

An object 5cm in length is placed at a distance of 20 cm in front of a concave mirror of radius of curvature 30 cm. Find the nature , position and size of object.

Given:

Height of object, $\sf h_o$

= 5 cm

Object distance, u = - 20 cm

Radius of curvature, R = 30 cm

Focal length, f = R/2 = 30/2 = - 15 cm

To find:

Position of image, it's nature and size?

Solution:

$\begin{gathered}\dag\;{\underline{\frak{Using\;mirror\;formula\;:}}}\\ \\\end{gathered}$

$\begin{gathered}\star\;{\boxed{\sf{\purple{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{- 15} = \dfrac{1}{v} + \dfrac{1}{- 20}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{1}{- 15} - \dfrac{1}{ - 20}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{1}{- 15} + \dfrac{1}{20}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{- 4 + 3}{60}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{- 1}{60}\\ \\\end{gathered}$

$\begin{gathered}:\implies{\boxed{\frak{\pink{v = - 60\;cm}}}}\;\bigstar\\ \\\end{gathered}$

$\begin{gathered}\therefore\;{\underline{\sf{Image\;distance\;is\; \bf{- 60\;cm}.}}}\\ \\\end{gathered}$

★ Nature of image:

Image is virtual and erect.

Image formed behind the mirror.

⠀━━━━━━━━━━━━━━━━━━━━━━━━━

Size of image,

$\begin{gathered}\dag\;{\underline{\frak{Using\; Formula\;of\;magnification\;:}}}\\ \\\end{gathered}$

$\begin{gathered}\star\;{\boxed{\sf{\purple{ \dfrac{h_i}{h_o} = \dfrac{v}{u}}}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{h_i}{5} = \cancel{ \dfrac{- 60}{20}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf h_i = - 3 \times 5\\ \\\end{gathered}$

$\begin{gathered}:\implies{\boxed{\frak{\pink{h_i = - 15\;cm}}}}\;\bigstar\\ \\\end{gathered}$

$\therefore\;{\underline{\sf{Height\;or\;size\;of\;image\;is\; \bf{- 15\;cm}.}}}$