Question

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Answer 1

Answer:

1.

In Concave mirror f and u are of same signs .

For erect image magnification must be positive ,

so

$m = \frac{v}{u} > 0 \\ \\ u > 0 \implies \: v > 0$

Use mirror formula

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} > 0 \\ \\ \frac{1}{12} - \frac{1}{u} > 0 \\ \\ \frac{1}{u} < \frac{1}{12} \\ \\ u > 12 \implies \: u \in \: ( \infty \: \: \: \: 12)$

$m = \frac{v}{u} = \frac{ \frac{1}{f} - \frac{1}{u} }{u} = \frac{u - 12}{ 12{u}^{2}} \\ \\ since \: u > 12 \: \: \\ {u}^{2} > u \implies \: 12 {u}^{2} > u \: \\ \implies \: 12 {u}^{2} + 12 > u \\ \\ \implies \frac{u - 12}{12 {u}^{2} } < 1 \\ \\ \implies \: m < 1$

So image will be smaller

Use simple lens formula for 3rd one

Answer 2

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