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

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:

$U = -10 cm$

$f = 15 cm$

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$So, \frac{1}{v} = \frac {1}{f} - \frac {1}{u}$

$\frac{1}{v} = \frac{1}{15} + \frac{1}{10}$

$\frac{1}{v} = \frac{(2+3)}{30} (lcm)$

$Therefore \: v = \frac{5}{30} = 6 cm$

Hence the image is formed behind the mirror and therefore it is virtual and erect.

Answer 2

${\bold{\orange{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}$

• $\rm\:Object\:distance\:(u)=-10\:cm$

• $\rm\:Focal\:length\:(f)=15\:cm$

${\bold{\pink{\underline{\pink{To}\:\purple{Fi}\blue{nd}\red{:-}}}}}$

• $\rm\:Image\:distance\:(v)=?$

${\bold{\purple{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}$

Using mirror formula:-

✭ $\pink{\underline{\boxed{\bf{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}}}}$

$\rm\longrightarrow\dfrac{1}{v}+\dfrac{1}{-10}=\dfrac{1}{15}$

$\rm\longrightarrow\dfrac{1}{v}-\dfrac{1}{10}=\dfrac{1}{15}$

$\rm\longrightarrow\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{10}$

$\rm\longrightarrow\dfrac{1}{v}=\dfrac{2+3}{30}$

$\rm\longrightarrow\dfrac{1}{v}=\cancel\dfrac{5}{30}$

$\rm\longrightarrow\dfrac{1}{v}=\dfrac{1}{6}$

$\rm\longrightarrow\:v=6\:cm$

Thus,the position of image is at distance of 6 cm from the convex mirror.

and nature of image is virtual and erect.