Question

An object is placed at a distance of 20 cm. in front of a convex lens whose radius of curvature is 40 cm. The image is formed at

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• Object is placed at Distance (u) = 20cm
• Radius of Curvature (R) = 40cm

$\color{darkblue}\underline{\underline{\sf To\:Find-}}$

• Image distance (v)

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We know that -

$\implies{\sf R=2f}$

$\implies{\sf f=\dfrac{R}{2}}$

$\implies{\sf f=\dfrac{40}{2} }$

$\color{orange}\implies{\sf f=+20\:cm }$

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For Convex Lens

• f = +ve
• u = -ve

━━━━━━━━━━━━━━━━

➩ Lens Formula

$\color{violet}\blacksquare{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}$

$\implies{\sf \dfrac{1}{20}=\dfrac{1}{v}-\left(-\dfrac{1}{20}\right) }$

$\implies{\sf \dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{20}}$

$\implies{\sf \dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{20}}$

$\color{red}\implies{\sf v=\infty }$

$\color{darkblue}\underline{\sf Nature\:of\:image}$

• $\color{red}{\sf Virtual}$ image in front on lens
• $\color{red}{\sf erect}$
• $\color{red}{\sf Enlarged}$

Answer 2

Given :-

A convex lens in which,

• object distance (u) = -20cm
• radius of curvature = 40cm

_____________________________

We know that,

R = 2f

So,

40 = 2f

f = 40/2

•°• f = 20cm

____________________________

$lens\: formula \: = \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Substituting values in the formula,

$\rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} \\ \rightarrow \frac{1}{v } = \frac{1}{( - 20)} + \frac{1}{20} \\ \rightarrow \: \frac{1}{v} = - \frac{1}{20} + \frac{1}{20} \\ \rightarrow \: \frac{1}{v} = \frac{ - 20 + 20}{20(20)} \\ \rightarrow \: \frac{1}{v} = \frac{0}{400} \\ \rightarrow \: v \: = 0$

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•°• Image formed is at infinity.