A 2.0 ,cm object is placed 15 ,cm in front of | vedclass

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(A) Given: Object height $h_o = 2.0 \,cm$, Object distance $u = -15 \,cm$, Focal length $f = -10 \,cm$. Using the mirror formula: $\frac{1}{v} + \frac

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$4 \,cm$, real

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(A) Given: Object height $h_o = 2.0 \,cm$, Object distance $u = -15 \,cm$, Focal length $f = -10 \,cm$. Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. Substituting the values: $\frac{1}{v} + \frac{1}{-15} = \frac{1}{-10} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} = -\frac{1}{30}$. So, $v = -30 \,cm$. The magnification $m = -\frac{v}{u} = -\frac{-30}{-15} = -2$. Image height $h_i = m 	imes h_o = -2 	imes 2.0 = -4.0 \,cm$. The negative sign indicates that the image is real and inverted. The size of the image is $4.0 \,cm$.

$A$ $2.0 \,cm$ object is placed $15 \,cm$ in front of a concave mirror of focal length $10 \,cm$. What is the size and nature of the image?

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