Question

An object is kept 30cm in front of a concave mirror of focal length 15cm find the position and nature of the image​

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf Position\:of\:image(v)=-30\:cm }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Object distance (u) = -30cm

• Focal lenght (f) = -15cm

$\large\underline\pink{\sf To\:Find: }$

• Position of image (v) = ?

• Nature of image .

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According to Mirror Formula :

$\large{\boxed{\sf \frac{1}{f}=\frac{1}{v}+\frac{1}{u}}}$

$\large\implies{\sf \dfrac{1}{-15}=\dfrac{1}{v}+\dfrac{1}{-30} }$

$\large\implies{\sf \dfrac{1}{-15}=\dfrac{1}{v}-\dfrac{1}{30} }$

$\large\implies{\sf \dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{30}}$

$\large\implies{\sf \dfrac{1}{v}=\dfrac{30+(-15)}{-15×30}}$

$\large\implies{\sf v=\dfrac{450}{15} }$

$\large\implies{\sf v=-30\:cm}$

$\red{\boxed{\sf Position\:of\:image(v)=-30\:cm }}$

Nature of Image (v) :-

• Image is formed at Centre of Curvature.

• Size of image is equal to size of object .

• Image formed is real and inverted.