Answer:
[tex]\boxed{\sf Position \ of \ the \ image = 6 \ cm}[/tex]
[tex]\boxed{\sf Nature \ of \ the \ image = Virtual \ and \ Erect}[/tex]
Given:
Object distance (u) = -10 cm
Focal length (f) = 15 cm
To Find:
Image distance (v) and nature of the image
Explanation:
Mirror Formula:
[tex]\boxed{\bold{\sf \frac{1}{v} + \frac{1}{u} = \frac{1}{f}}}[/tex]
Substituting values of u and f in the equation:
[tex]\sf \implies \frac{1}{v} -\frac{1}{10} =\frac{1}{15}[/tex]
[tex]\sf \implies \frac{1}{v}=\frac{1}{15} +\frac{1}{10}[/tex]
[tex]\sf \implies \frac{1}{v}=\frac{2}{30}+\frac{3}{30}[/tex]
[tex]\sf \implies \frac{1}{v}=\frac{2+3}{30}[/tex]
[tex]\sf \implies \frac{1}{v}=\frac{5}{30}[/tex]
[tex]\sf \implies v=\frac{30}{5}[/tex]
[tex]\sf \implies v=6 \ cm[/tex]
[tex]\therefore[/tex]
Image distance (v) = 6 cm
Nature of the image = Virtual and Erect
