Question

An object is placed at a distance of 10 cm from a convex mirror of focal length

15 cm. Find the position and nature of the image​

Answer 1

Given :-

Focal length of convex mirror = 15 cm

Object distance = -10 cm

To Find :-

The position and nature of the image​.

Analysis :-

Firstly, using the mirror formula you can get the position of the image accordingly.

Next find the position of the image accordingly.

Finally, using the formula of magnification you can find the size of the image formed easily.

Solution :-

We know that,

• f = Focal length
• u = Distance between mirror and object
• m = Magnification
• v = Distance between mirror and image

Using the formula,

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

Given that,

Focal length (f) = 15 cm

Object distance (u) = -10 cm

Substituting their values,

1/v = 1/15 - 1/-10

1/v = 2+3/30

1/v = 5/30

1/v = 6 cm

Therefore, the image is located at a distance of 6 cm from the mirror on the other side of the mirror.

Using the formula,

$\underline{\boxed{\sf Magnification=-\dfrac{Image \ distance}{Object \ distance} }}$

Given that,

Image distance (v) = -6 cm

Object distance (u) = -10 cm

Substituting their values,

m = -6/-10

m = 0.6

The positive and a value of less than 1 of magnification indicates that the image formed is virtual and erect and diminished.

Answer 2

$\bf \large ❥QUESTION❥$

An object is placed at a distance of 10 cm from a convex mirror of focal length

15 cm. Find the position and nature of the image

$\large\overline{\underline{ \boxed{ \sf \red{\bigstar \: formula \: used}}}}$

$\overline{\underline{ \boxed{ \sf \purple{\ \underline{\boxed{\sf Mirror \ formula=\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} }} : }}}}$

$\overline{\underline{ \boxed{ \sf \purple{ \: \:\underline{\boxed{\sf Magnification=-\dfrac{Image \ distance}{Object \ distance} }} }}}}$

$\bf \large ❥SOLUTION˙❥$

$\large{ \sf \underline{ \purple{we \: have : - }}}$

◕Focal length (f) = 15 cm

◕Object distance (u) = -10 cm

$\large{ \sf \underline{ \red{putting \: the \: values: - }}}$

by using mirror formula

$\bf \implies1/v = 1/15 - 1/-10 \\ </p><p></p><p> \bf \implies1/v = 2+3/30 \\ </p><p></p><p> \bf \implies1/v = 5/30 \\ </p><p></p><p> \bf \implies1/v = 6 cm</p><p>$

image distance =6cm

After that ,

by using magnification formula

$\large{ \sf \underline{ \purple{we \: have : - }}}$

✧Image distance (v) = -6 cm

✧Object distance (u) = -10 cm

$\large{ \sf \underline{ \red{putting \: the \: values: - }}}$

• m = -6/-10

• m = 0.6

┏ the image formed is virtual and erect and diminished.┛