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Physics

The image of a candle flame placed at a distance of 36 cm from a spherical lens, is formed on a screen placed at a distance of 72 cm from the lens. Calculate the focal length of the lens and its power.

Refraction Lens

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Answer

Given,

Object distance (u) = - 36 cm

As, image is formed on a screen then,

Image distance (v) = + 72 cm

Let, focal length be f.

From lens formula,

1f=1v−1u1f=172−1(−36)=172+136=1+2721f=372f=723=+24 cm\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \\[1 em] \dfrac{1}{\text f} = \dfrac{1}{72}-\dfrac{1}{(-36)}=\dfrac{1}{72}+\dfrac{1}{36}= \dfrac{1+2}{72} \\[1em] \dfrac{1}{\text f} = \dfrac{3}{72} \\[1 em] \text f = \dfrac{72}{3} = + 24 \text { cm}

Now,

Power=1f (in metre)=124×10−2=10024≈4.17 D\text {Power} = \dfrac{1}{\text {f (in metre)}} = \dfrac{1}{24\times 10^{-2}}=\dfrac{100}{24} ≈ 4.17\ \text D

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