Question

Help me please it's urgent
​

Answer 1

$\huge\bf{\red{\underline{\underline{\mathcal{AnSwer}}}}}$

Given:-

• Rear view mirror i.e it is a convex mirror
• Radius of curvature ( C ) = 3 m
• object distance ( u ) = -5 m

To find:-

• image distance ( v )
• nature of image
• size of image

Formula:-

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

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

Solution:-

$\implies$ $\sf\dfrac{1}{3}$ = $\sf\dfrac{1}{v}$ + $\sf\dfrac{1}{-5}$

$\implies$ $\sf\dfrac{1}{v}$ = $\sf\dfrac{1}{3}$ + $\sf\dfrac{1}{5}$

$\implies$ $\sf\dfrac{1}{v}$ = $\sf\dfrac{3+5}{15}$

$\implies$ $\sf\dfrac{1}{v}$ = $\sf\dfrac{8}{15}$

$\implies$ $\sf v$ = $\sf 1.25$

Nature:- The positive sign of this image shows that it is Virtual and erect image.

Size of this image

$\implies$ m = $\sf\dfrac{-1.25}{-5}$

$\implies$ m = 0.25

hence, this image is smaller than the object

_________________________

Additional information:-

Cases in Concave mirror -

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{cccc}\sf \pink{Position_{\:(object)}} &\sf \purple{Position_{\:(image)}} &\sf \red{Size_{\:(image)}} &\sf \blue{Nature_{\:(image)}}\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf At \:Infinity &\sf At\: F&\sf Highly\:Diminished&\sf Real\:and\:Inverted\\\\\sf Beyond\:C &\sf Between\:F\:and\:C&\sf Diminished&\sf Real\:and\:Inverted\\\\\sf At\:C &\sf At\:C&\sf Same\:Size&\sf Real\:and\:Inverted\\\\\sf Between\:C\:and\:F&\sf Beyond\:C&\sf Enlarged&\sf Real\:and\;Inverted\\\\\sf At\:F&\sf At\:Infinity&\sf Highly\: Enlarged&\sf Real\:and\:Inverted\\\\\sf Between\:F\:and\:P&\sf \: Behind\:the\:mirror&\sf Enlarged&\sf \: Erect\:and\:Virtual\end{array}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

ɢɪᴠᴇɴ:-

ʀᴇᴀʀ ᴠɪᴇᴡ ᴍɪʀʀᴏʀ ɪ.ᴇ ɪᴛ ɪs ᴀ ᴄᴏɴᴠᴇx ᴍɪʀʀᴏʀ

ʀᴀᴅɪᴜs ᴏғ ᴄᴜʀᴠᴀᴛᴜʀᴇ ( ᴄ ) = 3 ᴍ

ᴏʙᴊᴇᴄᴛ ᴅɪsᴛᴀɴᴄᴇ ( ᴜ ) = -5 ᴍ

ᴛᴏ ғɪɴᴅ:-

ɪᴍᴀɢᴇ ᴅɪsᴛᴀɴᴄᴇ ( ᴠ )

ɴᴀᴛᴜʀᴇ ᴏғ ɪᴍᴀɢᴇ

sɪᴢᴇ ᴏғ ɪᴍᴀɢᴇ

ғᴏʀᴍᴜʟᴀ:-

\ɪᴍᴘʟɪᴇs⟹ \ᴜɴᴅᴇʀʟɪɴᴇ{\ʙᴏxᴇᴅ{\sғ \ᴅғʀᴀᴄ{1}{ғ} = \ᴅғʀᴀᴄ{1}{ᴠ} + \ᴅғʀᴀᴄ{1}{ᴜ}}}

ғ

1

=

ᴠ

1

+

ᴜ

1

\ɪᴍᴘʟɪᴇs⟹ \ᴜɴᴅᴇʀʟɪɴᴇ{\ʙᴏxᴇᴅ{\sғ ᴍ = \ᴅғʀᴀᴄ{-ᴠ}{ᴜ}}}

ᴍ=

ᴜ

−ᴠ

sᴏʟᴜᴛɪᴏɴ:-

\ɪᴍᴘʟɪᴇs⟹ \sғ\ᴅғʀᴀᴄ{1}{3}

3

1

= \sғ\ᴅғʀᴀᴄ{1}{ᴠ}

ᴠ

1

+ \sғ\ᴅғʀᴀᴄ{1}{-5}

−5

1

\ɪᴍᴘʟɪᴇs⟹ \sғ\ᴅғʀᴀᴄ{1}{ᴠ}

ᴠ

1

= \sғ\ᴅғʀᴀᴄ{1}{3}

3

1

+ \sғ\ᴅғʀᴀᴄ{1}{5}

5

1

\ɪᴍᴘʟɪᴇs⟹ \sғ\ᴅғʀᴀᴄ{1}{ᴠ}

ᴠ

1

= \sғ\ᴅғʀᴀᴄ{3+5}{15}

15

3+5

\ɪᴍᴘʟɪᴇs⟹ \sғ\ᴅғʀᴀᴄ{1}{ᴠ}

ᴠ

1

= \sғ\ᴅғʀᴀᴄ{8}{15}

15

8

\ɪᴍᴘʟɪᴇs⟹ \sғ ᴠᴠ = \sғ 1.251.25

ɴᴀᴛᴜʀᴇ:- ᴛʜᴇ ᴘᴏsɪᴛɪᴠᴇ sɪɢɴ ᴏғ ᴛʜɪs ɪᴍᴀɢᴇ sʜᴏᴡs ᴛʜᴀᴛ ɪᴛ ɪs ᴠɪʀᴛᴜᴀʟ ᴀɴᴅ ᴇʀᴇᴄᴛ ɪᴍᴀɢᴇ.

sɪᴢᴇ ᴏғ ᴛʜɪs ɪᴍᴀɢᴇ

\ɪᴍᴘʟɪᴇs⟹ ᴍ = \sғ\ᴅғʀᴀᴄ{-1.25}{-5}

−5

−1.25

\ɪᴍᴘʟɪᴇs⟹ ᴍ = 0.25

ʜᴇɴᴄᴇ, ᴛʜɪs ɪᴍᴀɢᴇ ɪs sᴍᴀʟʟᴇʀ ᴛʜᴀɴ ᴛʜᴇ ᴏʙᴊᴇᴄᴛ

_________________________

ᴀᴅᴅɪᴛɪᴏɴᴀʟ ɪɴғᴏʀᴍᴀᴛɪᴏɴ:-

ᴄᴀsᴇs ɪɴ ᴄᴏɴᴄᴀᴠᴇ ᴍɪʀʀᴏʀ -

\ʙᴇɢɪɴ{ɢᴀᴛʜᴇʀᴇᴅ}\ʙᴇɢɪɴ{ɢᴀᴛʜᴇʀᴇᴅ}\ʙᴇɢɪɴ{ɢᴀᴛʜᴇʀᴇᴅ}\ʙᴇɢɪɴ{ɢᴀᴛʜᴇʀᴇᴅ}\ʙᴇɢɪɴ{ɢᴀᴛʜᴇʀᴇᴅ}\ʙᴏxᴇᴅ{\ʙᴇɢɪɴ{ᴀʀʀᴀʏ}{ᴄᴄᴄᴄ}\sғ \ᴘɪɴᴋ{ᴘᴏsɪᴛɪᴏɴ_{\:(ᴏʙᴊᴇᴄᴛ)}} &\sғ \ᴘᴜʀᴘʟᴇ{ᴘᴏsɪᴛɪᴏɴ_{\:(ɪᴍᴀɢᴇ)}} &\sғ \ʀᴇᴅ{sɪᴢᴇ_{\:(ɪᴍᴀɢᴇ)}} &\sғ \ʙʟᴜᴇ{ɴᴀᴛᴜʀᴇ_{\:(ɪᴍᴀɢᴇ)}}\\\ғʀᴀᴄ{\ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ}{}&\ғʀᴀᴄ{\ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ}{}&\ғʀᴀᴄ{\ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ\ǫǫᴜᴀᴅ}{}&\ғʀᴀᴄ{\ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ \ǫǫᴜᴀᴅ\ǫǫᴜᴀᴅ}{}\\\sғ ᴀᴛ \:ɪɴғɪɴɪᴛʏ &\sғ ᴀᴛ\: ғ&\sғ ʜɪɢʜʟʏ\:ᴅɪᴍɪɴɪsʜᴇᴅ&\sғ ʀᴇᴀʟ\:ᴀɴᴅ\:ɪɴᴠᴇʀᴛᴇᴅ\\\\\sғ ʙᴇʏᴏɴᴅ\:ᴄ &\sғ ʙᴇᴛᴡᴇᴇɴ\:ғ\:ᴀɴᴅ\:ᴄ&\sғ ᴅɪᴍɪɴɪsʜᴇᴅ&\sғ ʀᴇᴀʟ\:ᴀɴᴅ\:ɪɴᴠᴇʀᴛᴇᴅ\\\\\sғ ᴀᴛ\:ᴄ &\sғ ᴀᴛ\:ᴄ&\sғ sᴀᴍᴇ\:sɪᴢᴇ&\sғ ʀᴇᴀʟ\:ᴀɴᴅ\:ɪɴᴠᴇʀᴛᴇᴅ\\\\\sғ ʙᴇᴛᴡᴇᴇɴ\:ᴄ\:ᴀɴᴅ\:ғ&\sғ ʙᴇʏᴏɴᴅ\:ᴄ&\sғ ᴇɴʟᴀʀɢᴇᴅ&\sғ ʀᴇᴀʟ\:ᴀɴᴅ\;ɪɴᴠᴇʀᴛᴇᴅ\\\\\sғ ᴀᴛ\:ғ&\sғ ᴀᴛ\:ɪɴғɪɴɪᴛʏ&\sғ ʜɪɢʜʟʏ\: ᴇɴʟᴀʀɢᴇᴅ&\sғ ʀᴇᴀʟ\:ᴀɴᴅ\:ɪɴᴠᴇʀᴛᴇᴅ\\\\\sғ ʙᴇᴛᴡᴇᴇɴ\:ғ\:ᴀɴᴅ\:ᴘ&\sғ \: ʙᴇʜɪɴᴅ\:ᴛʜᴇ\:ᴍɪʀʀᴏʀ&\sғ ᴇɴʟᴀʀɢᴇᴅ&\sғ \: ᴇʀᴇᴄᴛ\:ᴀɴᴅ\:ᴠɪʀᴛᴜᴀʟ\ᴇɴᴅ{ᴀʀʀᴀʏ}}\ᴇɴᴅ{ɢᴀᴛʜᴇʀᴇᴅ}\ᴇɴᴅ{ɢᴀᴛʜᴇʀᴇᴅ}\ᴇɴᴅ{ɢᴀᴛʜᴇʀᴇᴅ}\ᴇɴᴅ{ɢᴀᴛʜᴇʀᴇᴅ}\ᴇɴᴅ{ɢᴀᴛʜᴇʀᴇᴅ}

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