Question

Help!!! Answer question b) using thin lens equations PLEASE

Answer 1

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Given:

• Distance of object(u) = -32cm
• Focal length(f) = +8cm
• Height of object(h1) = +3cm

To Find:

• Image location (distance of image from lens)(v)
• Height of image(h2)

Solution:

$\bf{1.\:\: Image\: location} \\\\\text{By Lens formula:} \\\\:\implies \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\\\:\implies \dfrac{1}{v} - \dfrac{1}{-32} = \dfrac{1}{8} \\\\:\implies \dfrac{1}{v} + \dfrac{1}{32} = \dfrac{1}{8} \\\\:\implies \dfrac{1}{v} = \dfrac{1}{8} - \dfrac{1}{32} \\\\:\implies \dfrac{1}{v} = \dfrac{4}{32} - \dfrac{1}{32} \\\\:\implies \dfrac{1}{v} = \dfrac{3}{32} \\\\\bf{:\implies v = \dfrac{32}{3} \approx +10.67cm}$

$\bf{2. \:\:Height \:of \:image} \\\\\text{By Magnification Formula:} \\\\:\implies m = \dfrac{v}{u} = \dfrac{h_2}{h_1} \\\\:\implies \dfrac{v}{u} = \dfrac{h_2}{h_1} \\\\:\implies \dfrac{\dfrac{+32\!\!\!\!\!{/}}{3}}{-32\!\!\!\!\!{/}} = \dfrac{h_2}{+3} \\\\:\implies \dfrac{-1}{3\!\!\!{/}} = \dfrac{h_2}{3\!\!\!{/}} \\\\\bf{:\implies h_2 = -1cm}$

Formulae used:

• Lens formula => 1/v - 1/u = 1/f
• Magnification formula => m = v/u = h2/h1

Learn More:

• Mirror formula => 1/v + 1/u = 1/f
• Magnification formula => m = -v/u = h2/h1

Answer 2

Given:

• Distance of object(u) = -32cm
• Focal length(f) = +8cm
• Height of object(h1) = +3cm

To Find:

• Image location (distance of image from lens)(v)
• Height of image(h2)

Answer:

1. ɪᴍᴀɢᴇʟᴏᴄᴀᴛɪᴏɴ

ʙʏ ʟᴇɴs ғᴏʀᴍᴜʟᴀ

• Lens formula => 1/v - 1/u = 1/f

$⟹ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$⟹ \frac{1}{v} - \frac{1}{ - 32} = \frac{1}{8}$

$⟹ \frac{1}{v} + \frac{1}{32} = \frac{1}{8}$

$\frac{1}{v} = \frac{1}{8} - \frac{1}{32}$

$\frac{1}{v} = \frac{4}{32} - \frac{1}{32}$

$\frac{1}{v} = \frac{3}{32}$

$v = \frac{32}{3} ≈ + 10.67 \: cm$

2.ʜᴇɪɢʜᴛ ᴏғ ɪᴍᴀɢᴇ

ʙʏ ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ ғᴏʀᴍᴜʟᴀ

• Magnification formula => m = v/u = h2/h1

$⟹ \: m = \frac{v}{u} = \frac{h_{2} }{h_{1} }$

$⟹ \frac{v}{u} = \frac{h_{2} }{h_{1} }$

$⟹ \frac{ \frac{ + 32}{3} }{ - 32} = \frac{h _{2} }{ + 3}$

$⟹ \frac{ - 1}{3} = \frac{h_{2} }{3}$

$h_{2} = - 1cm$