Question

an object of height 4 cm is placed at a distance of 15 CM in front of concave lens of power minus 10 diopter find the size of image please give correct answer ​

Answer 1

Answer:

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Explanation:

$to \: find = \\ size \: of \: image \: (h2) \\ \\ so \: for \: a \: concave \: lens \\ height \: of \: object \: (h1) = 4.cm \\ object \: distance \: (u) = - 15.cm \\ power \: of \: lens \: (P) = - 10.D \\ \\ so \: we \: know \: that \\ power \: (P) = \frac{1}{f \: (m)} \\ \\ hence \: \\ - 10 = \frac{1}{f \: (m)} \\ \\ ie \: f = \frac{1}{ - 10} \\ \\ = - 0.1 \: m \\ = - 0.1 \times 100 \\ = - 10.cm \\ \\ hence \: then \\ \: focal \: length \: (f) = - 10.cm$

$now \: by \: applying \: lens \: formula \\ we \: get \\ \\ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \\ \frac{1}{v} - \frac{1}{( - 15)} = \frac{1}{ (- 10)} \\ \\ \frac{1}{v} + \frac{1}{15} = \frac{ - 1}{10} \\ \\ \frac{1}{v} = \frac{ -1 }{10} - \frac{1}{15} \\ \\ = \frac{ - 15 - 10}{10 \times 15} \\ \\ = \frac{ - 25}{150} \\ \\ = \frac{ - 1}{6} \\ \\ thus \: applying \: invertendo \\ we \: get \\ v = - 6.cm \\ hence \: image \: is \: virtual \: and \: erect \: and \: smaller \: than \: object$

$now \: applying \\ \\ \frac{h2}{h1} = \frac{v}{u} \\ \\ \frac{h2}{4} = \frac{ - 6}{ - 15} \\ \\ \frac{h2}{4} = \frac{6}{15} \\ \\ \: h2 = \frac{6 \times 4}{15} \\ \\ = \frac{24}{15} \\ \\ = 1.6 \\ \\ \\ hence \: then \\ \: height \: of \: image \: (h2) = 1.6 \: cm \\ \\ thus \: size \: of \: image \: is \: 1.6 \: cm$