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Confusion in sign convention of magnification?

0

$begingroup$

A concave mirror forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of object. If the shift of object is 6 cm. Find the focal length and initial position of object?

My attempt :-

1st way

As per question
$$m_i=3$$
$$Longrightarrow frac-f-f+u=3$$
$$Longrightarrow 2f=3u. ...(1)$$
For second case , On a similar way we get
$$ frac-f-f+(u+6)=3$$
$$ Longrightarrow f=2u+12. ....(2)$$
On solving equation 1 and 2 we get
$$f=-36 , u=-24$$

2nd way

Since the image is inverted the magnification must be negative.

So $$m_i =-3$$
$$Longrightarrow fracff-u=-3$$
For second case
$$fracff-(u+6)=-2$$
On solving we get
$$f=-36, u=-48$$

My question:- why I get two different answer, if use the sign. Convention in the formula of magnification I get different results when I use the sign. Convention in the value of magnification. Why this this happen and which is right process ?

physics

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asked Apr 8 at 6:43

Abhishek KumarAbhishek Kumar

685

New contributor

Abhishek Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

A concave mirror forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of object. If the shift of object is 6 cm. Find the focal length and initial position of object?

My attempt :-

1st way

As per question
$$m_i=3$$
$$Longrightarrow frac-f-f+u=3$$
$$Longrightarrow 2f=3u. ...(1)$$
For second case , On a similar way we get
$$ frac-f-f+(u+6)=3$$
$$ Longrightarrow f=2u+12. ....(2)$$
On solving equation 1 and 2 we get
$$f=-36 , u=-24$$

2nd way

Since the image is inverted the magnification must be negative.

So $$m_i =-3$$
$$Longrightarrow fracff-u=-3$$
For second case
$$fracff-(u+6)=-2$$
On solving we get
$$f=-36, u=-48$$

My question:- why I get two different answer, if use the sign. Convention in the formula of magnification I get different results when I use the sign. Convention in the value of magnification. Why this this happen and which is right process ?

physics

share|cite|improve this question

asked Apr 8 at 6:43

Abhishek KumarAbhishek Kumar

685

New contributor

Abhishek Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

A concave mirror forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of object. If the shift of object is 6 cm. Find the focal length and initial position of object?

My attempt :-

1st way

As per question
$$m_i=3$$
$$Longrightarrow frac-f-f+u=3$$
$$Longrightarrow 2f=3u. ...(1)$$
For second case , On a similar way we get
$$ frac-f-f+(u+6)=3$$
$$ Longrightarrow f=2u+12. ....(2)$$
On solving equation 1 and 2 we get
$$f=-36 , u=-24$$

2nd way

Since the image is inverted the magnification must be negative.

So $$m_i =-3$$
$$Longrightarrow fracff-u=-3$$
For second case
$$fracff-(u+6)=-2$$
On solving we get
$$f=-36, u=-48$$

My question:- why I get two different answer, if use the sign. Convention in the formula of magnification I get different results when I use the sign. Convention in the value of magnification. Why this this happen and which is right process ?

physics

share|cite|improve this question

asked Apr 8 at 6:43

Abhishek KumarAbhishek Kumar

685

New contributor

Abhishek Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Apr 8 at 6:43

Abhishek KumarAbhishek Kumar

685

New contributor

Abhishek Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Apr 8 at 6:43

Abhishek KumarAbhishek Kumar

685

New contributor

Abhishek Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 8 at 6:43

Abhishek KumarAbhishek Kumar

685

New contributor

Abhishek Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 8 at 6:43

Abhishek KumarAbhishek Kumar

685