Question

$\large\red{\underline{{\boxed{\textbf{question:-}}}}}$
A converging lens of refractive index 1.5 is kept in a liquid medium having same refractive index. What would be the focal length of the lens in this medium?


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Answer 1

We have, Len's maker formula =

f

1

=(

μ

1

μ

2

−1)(

R

1

1

−

R

2

1

),

Here, μ

1

=μ

2

=1.5

⇒

f

1

=0

⇒f=∞

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Answer 2

$\frak{Given}\:\begin{cases}\:\quad \sf The\:Refractive \:Index\:of\:Lens \:is\:\pmb{\frak{1.5}}\\ \:\quad \sf Refractive \:Index\:_{\:(Lens)}\:=\:Refractive \:Index\:_{\:(Medium)}\:\end{cases}$

Need To Find : Focal Length of the Lens in medium ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \bigstar \pmb{\underline {\sf \: From \: Lens \:Maker \:Formula \: \::\:}}$

$\qquad \star\:\pmb{\underline {\boxed {\pink{\sf \: \dfrac{1}{f}\:=\:\bigg( \dfrac{\mu_{material\:}}{\mu_{medium\:}} \:-\:1\bigg) \:\bigg\lgroup \dfrac{1}{R_1}\: -\:\dfrac{1}{R_2}\:\bigg\rgroup}}}}$

$\frak{Where}\:\begin{cases}\:\quad \sf Refractive \:Index_{\:(Material)\:} \:,\: \mu_{material} \:=\:\pmb{\frak{1.5}}\\ \:\quad \sf Refractive \:Index_{\:(Medium)\:} \:,\: \mu_{medium} \:=\:\pmb{\frak{1.5}}\:\end{cases}$

$\\\qquad \dag\underline {\pmb{\frak{ Substituting \:known \:Values \:in \:Given \:Formula \:\::\:}}}$

$\twoheadrightarrow \sf \: \dfrac{1}{f}\:=\:\bigg( \dfrac{\mu_{material\:}}{\mu_{medium\:}} - 1\:\bigg) \:\bigg\lgroup \dfrac{1}{R_1}\: -\:\dfrac{1}{R_2}\:\bigg\rgroup\\\\\\ \twoheadrightarrow \sf \: \dfrac{1}{f}\:=\:\bigg( \dfrac{1.5}{1.5} - 1\:\bigg) \:\bigg\lgroup \dfrac{1}{R_1}\: -\:\dfrac{1}{R_2}\:\bigg\rgroup\\\\\\ \twoheadrightarrow \sf \: \dfrac{1}{f}\:=\:\bigg( 1 - 1\:\bigg) \\\\\\\twoheadrightarrow \sf \: \dfrac{1}{f}\:=\:1 - 1 \\\\\\\twoheadrightarrow \sf \: \dfrac{1}{f}\:=\:0 \\\\\\\twoheadrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:f\:\:=\:\infty\:}}}}}\:\bigstar \:$

$\:\:\therefore \:\underline {\sf Hence, \:Focal\:Length \:,f\:of \:of \:Lens\:is\:\pmb{\sf \infty}\:.}$