Question

HOLA,





$\huge \fbox \red{❥ Question}$




A concave lens has focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.​

Answer 1

here is your answer full and correct answer

please mark as brainliest answer

Answer 2

$\color{orange} \huge \underline{ \underline{\cal GIVEN}}$

As the lens used is concave so it's focal length will be negative

• Focal length,F=20cm
• Height of object,H=5cm
• Distance of image,V=15cm

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$\color{green} \huge \cal \underline{\underline{TO \: FIND} }$

• Distance of object from lens,u
• Height of the image

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$\color{blue} \huge \cal \underline{ \underline{SOLUTION}}$

First we apply lens formula to find distance of object

$\color{navy}\LARGE \sf \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\LARGE \sf \to\frac{1}{20} = \frac{1}{15} - \frac{1}{u}$

$\LARGE \sf \to \frac{1}{u} = \frac{1}{15} - \frac{1}{20}$

$\LARGE \sf \to \frac{1}{u} = \frac{1}{60}$

$\LARGE \sf \color{grey} Cross \: multiply$

$\LARGE \boxed{ \sf \color{gold}\to {u} = {60} cm}$

So distance of object from the mirror is 60m.

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Now we have to find size of image.

Now we apply formula of magnification:

$\color{black} \LARGE \sf M= \frac{v}{u}......(1)$

$\color{black} \LARGE \sf \: M= \frac{Height \: of \: image}{Height \: of \: object}....(2)$

$\LARGE \sf From \: (1) \: and \: (2)$

$\color{red} \LARGE \sf \frac{v}{u} = \frac{Height \: of \: image}{Height \: of \: object}$

$\LARGE \sf Put \: the \: given \: values$

$\color{pink} \LARGE \sf \frac{15}{60} = \frac{Height \: of \: image}{5}$

$\LARGE \sf \frac{15 \times 5}{60} = {Height \: of \: image}{}$

$\LARGE \color{lightgreen}\sf 1.5cm = {Height \: of \: image}{}$