Math, asked by sanjanajirankali, 9 months ago

8. If the object is placed 6 cm from a concave mirror, the image distance is 12 cm. What is the
image distance if the object is moved from the original position 1 cm away from the mirror?

Answers

Answered by ıtʑFᴇᴇʟɓᴇãᴛ
27

\mathcal{\huge{\fbox{\red{Question:-}}}}

✴ If the object is placed 6 cm from a concave mirror, the image distance is 12 cm. What is the image distance if the object is moved from the original position 1 cm away from the mirror?

\mathcal{\huge{\fbox{\green{AnSwEr:-}}}}

The image distance is -9.3cm.

\mathcal{\huge{\fbox{\purple{Solution:-}}}}

Given :-

  • If the object is placed 6 cm from a concave mirror, the image distance is 12 cm.

  • The original position of object is 1 cm away from the mirror.

To Find :-

  • The image distance if the object is moved from the original position 1 cm away from the mirror.

Calculation :-

We have,

  • Object distance (u) = - 6cm

  • Image distance (v) = - 12cm

  • Focal length (f) = ?

Applying mirror Formula here,

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

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

 \dfrac{1}{f}  =  \dfrac{1}{ - 12}  +  \dfrac{1}{ - 6}

 \dfrac{1}{f}  =  \dfrac{ - 1 \times - 2}{12}

 \dfrac{1}{f}  =  \dfrac{ - 3}{12}

 \dfrac{1}{f}  =  -  \dfrac{1}{4}

f \:  =  - 4 \: cm

Now, we have our focal length -4 cm .

According to the question,

if the object is moved from the original position 1 cm away from the mirror Then,

  • New, Object distance = -(6 + 1) = -7

  • Focal length (f) = -4

We have to find our Image distance (v).

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

 \dfrac{1}{v}  =  \dfrac{1}{f}   -  \dfrac{1}{u}

 \dfrac{1}{v}  =  \dfrac{1}{ - 4}   -   \dfrac{1}{ - 7}

 \dfrac{1}{v}  =  \dfrac{ - 1}{ 4}  +  \dfrac{  1}{  7}

 \dfrac{1}{v}  =   \dfrac{ - 7 + 4}{28}

 \dfrac{1}{v}  =   \dfrac{ - 3}{28}

 - 3v \:  = 28

v =  \cancel{\dfrac{28}{ - 3}}

➡ v = - 9.3cm

Hence,our Image distance (v) is 9.3 cm, if we keep the object 1 cm away from the mirror.

___________________________________


Anonymous: Great :)
Answered by Anonymous
16

\large\bf\underline \blue {To \: \mathscr{f}ind:-}

  • we need to find image distance if the object is moved from the original position 1 cm away from the mirror.

 \huge\bf\underline \red{ \mathcal{S}olution:-}

 \bf\underline{\purple{Given:-}}

Case (i):-

If the image is real in front of the mirror , then the image distance is negative.

So,

object distance (u) = - 6cm

image distance (v) = - 12cm

Focal length (f) = ?

By using mirror Formula:-

 \star \: \large \green{ \bf \: \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }

↛ 1/f = 1/ -12 + 1/-6

↛ 1/f = -1 - 2/12

↛ 1/f = -3/12

↛ 1/ f = -1/4

↛ - f = 4 f = - 4cm

Now,

 \underline{ \large \purple{ \mathscr{\dag\:A \bf{ccourding} \: to \: \mathscr {Q} \bf{uestion} ....}}}

if the object is moved from the original position 1 cm away from the mirror .

Then,

  • Object distance = -(6 + 1) = -7cm
  • Focal length (f) = -4
  • image distance (v) = ?

↛1/f = 1/v + 1/u

↛ 1/v = 1/f - 1/u

→ 1/v = 1/-4 -1/-7

→ 1/v = -1/4 + 1/7

→ 1/v = (-7 + 4)/28

→ 1/v = -3/28

→ -3v = 28

→ v = 28/-3

  • v = - 9.3cm

Hence,

  • image distance will be -9.3cm if the object is moved from the original position 1 cm away from the mirror.

━━━━━━━━━━━━━━━━━━━━━━━━━

Case (ii) :-

If the image is virtual behind the mirror, then image distance is positive,

So,

  • object distance (u) = - 6cm
  • image distance (v) = 12cm
  • Focal length (f) = ?

  • By using mirror Formula:-

 \star \: \large \green{ \bf \: \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }

↛ 1/f = 1/ 12 + 1/-6

↛ 1/f = (1 - 2)/12

↛ 1/f = -1/12

↛ 1/ f = -1/12

↛ - f = 12

  • f = -12cm

 \underline{ \large \purple{ \mathscr{\dag\:A \bf{ccourding} \: to \: \mathscr {Q} \bf{uestion} ....}}}

if the object is moved from the original position 1 cm away from the mirror

Then,

  • Object distance = -(6 + 1) = -7
  • Focal length (f) = -12cm
  • image distance (v) = ?

↛1/f = 1/v + 1/u

↛ 1/v = 1/f - 1/u

→ 1/v = 1/-12 -1/-7

→ 1/v = -1/12 + 1/7

→ 1/v = (-7 + 12)/84

→ 1/v = 5/84

→ 5v = 84

→ v = 84/5 v = 16.8cm

Hence,

  • image distance will be 16.8cm if the object is moved from the original position 1 cm away from the mirror.

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