Question

In object 2 cm high is placed at a distance of 64 cm from a white screen. On placing a convex lens at a distance of 32 cm from the object it is found that a distinct image of the object is formed on the screen. What is the focal length of the convex lens and size of the image formed on the screen? Draw a ray diagram to show the formation of the image in this position of the object with respect to the lens.​

Answer 1

Hey friend :)

Here's your answer :-

$\: Given \: \: data→$

$Height /size\: of \: an \: object \: , h _{1} = 2 \: cm \\ Distance \: of \: Object, u = - 32cm \\ Image \: distance \:, v = 64 - 32 = 32cm \: ...(Real \: image)$

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$\: To \: find→$

$Height/size \: \: of \: \: image,h _{2} = ? \\ Focal \: \: length ,f= ?$

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$Solution \: →$

$\: \: \: \: \: \: \: We \: \: know,$

$\boxed{\sf\purple{ \frac{1}{f} ~=~ \frac{1}{v} ~ - ~ \frac{1}{u} }} \: ┈┈┈(Lens \: \: formula)$

$⛬ \: \: \: \frac{1}{f} = \frac{1}{32} - (\frac{1}{ - 32} ) \\ \\ ↦ \frac{1}{f} = \frac{1}{32} + \frac{1}{32} \\ \\ ↦ \frac{1}{f} = \frac{1 + 1}{32} \\ \\↦ \frac{1}{f} = \frac{2}{32} \\ \\ ↦\frac{1}{f} = \frac{1}{16} \\ \\ \sf \underline\mathtt\blue{↦ \: f = 16cm}$

$A/c \: \: To \: \: magnification \: \: formula \: →$

$→ \boxed{\sf\purple{\: \frac{h _{2}}{h _{1} } = \frac{v}{u} }}$

$→ \frac{h _{2} }{2} = (\frac{32}{ - 32} ) \\ \\ → \frac{h _{2} }{2} = \frac{32}{ - 32} \\ \\ → \frac{h _{2} }{2} = - 1 \\ \\ → h _{2} = - 1 \times 2 \\ \\ → \underline\mathtt\blue{h _{2} = - 2cm } \: \: \: \: \:$

$So, \: The \: size \: of \: the \: object \: = \: h _{2} = \: -2cm \\ \\</p><p>The \: Focal \: length \: = f = 16 cm </p><p>$

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@llSugarXSweetll