Question

An object 4cm in height,is placed at 15 cm in front of a concave mirror of foacl length 10cm. At what distance from mirror should acreen be placef

Answer 1

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Answer 2

$\large{\red{\bold{\underline{Given:}}}}$

$\sf \: (i) \: h_{1} = + 4cm \\ \sf \: (ii) \: f = - 10cm \\ \sf \: (iii) \: u = - 15cm$

$\large{\green{\bold{\underline{To \: Find:}}}}$

$\sf \: v = ????$

$\large{\blue{\bold{\underline{Mirror \: Formula:}}}}$

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

$\large{\red{\bold{\underline{Calculation:}}}}$

$\rightarrow \: \sf \: \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \\ \rightarrow \: \sf \: \frac{1}{v} = - \frac{1}{10} - ( - \frac{1}{15}) \\ \\ \rightarrow \: \sf \: \frac{1}{v} = \frac{ - 1 \times 3 + 1 \times 2}{30} \\ \\ \rightarrow \: \sf \: \frac{1}{v} = \frac{ - 3 + 2}{30} \\ \\ \rightarrow \: \sf \: \frac{1}{v} = \frac{ - 1}{30} \\ \\ \rightarrow \: \sf \: \large{ \boxed{v = - 30cm}}$

$\large{\orange{\bold{\underline{Extra \: Information:}}}}$

We can also calculation height of the Image

$\rightarrow \: \sf \: \frac{h_{1}}{h_{2}} = - \frac{v}{u} \\ \\ \rightarrow \: \sf \: h_{2} = - \frac{v}{u} \times h_{1} \\ \\ \rightarrow \: \sf \: h_{2} = \frac{ - 30}{ - 15} \times 4 \\ \\ \rightarrow \: \sf \: h_{2} = \frac{ \cancel- 30}{ \cancel - 15} \times 4 \\ \\ \rightarrow \: \sf \: h_{2} = 2 \times 4 \\ \\ \rightarrow \: \sf \: \boxed{h_{2} = 8cm}$