Question

find the refractive index of the glass which is a symmetrical convex lens if its focal length is equal to the radius of curvature of its surface

Answer 1

$\rm \huge \underline\mathfrak\green{Given:-}$

$\rm\dashrightarrow{ R_{(1)} = R = f}$

$\rm\dashrightarrow{ R_{(2)} = R = f}$

$\rm\huge \underline\mathfrak \green{Find :-}$

• find the refractive index of the glass.

$\rm\huge\underline\mathfrak \green{ Solution:-}$

We know that,

$\rm \dashrightarrow{Refractive \: index \: of \: glass = 1.5.}$

$\: \: \: \longrightarrow\rm{ \frac{1}{f} = (n - 1) \times \frac{2}{f} }$

$\: \: \: \: \longrightarrow\rm{ 2n - 2 = 1}$

$\: \: \: \: \: \: \longrightarrow\rm{ 2n = 3}$

$\: \: \: \: \: \: \longrightarrow\rm{ n = \frac{3}{2} }$

$\: \: \: \: \: \: \rm\longrightarrow \fbox \red{n = 1.5}$

━─━─━─━─━─━─━─━─━─━─━─━

$\bf\bigstar\green { \: Physics \: \: formula:- }$

$\: \: \: \: \: \: \begin{gathered}\boxed{\begin{array}{cc}\sf formula&\\\frac{\qquad }{ }{}\\\sf\frac{1}{f} = \frac{1}{v} - \frac{1}{u} &\ \\\\\sf \: v = v_{o} + at &\\\\\sf p = mv &\\\\\sf \: p = f + v&\\ &\end{array}}\end{gathered}$

Answer 2

Answer:

$\underline{ \boxed{\mathbb{\red{⛄solution♡࿐⛄}}}} \\ \\ We \: know \: that, \\ </p><p></p><p>\rm \dashrightarrow{Refractive \: index \: of \: glass = 1.5 . } \\ \\ ⇢Refractive \: index \: of \: glass=1.5. \\ </p><p></p><p>\: \: \: \longrightarrow\rm{ \frac{1}{f} = (n - 1) \times \frac{2}{f} } \\ ⟶f1=(n−1)×f2 \\ </p><p></p><p>\: \: \: \: \longrightarrow\rm{ 2n - 2 = 1} \\ ⟶2n−2=1 \\ </p><p></p><p>\: \: \: \: \: \: \longrightarrow\rm{ 2n = 3} \\ ⟶2n=3 \\ </p><p></p><p>\: \: \: \: \: \: \longrightarrow\rm{ n = \frac{3}{2} } \\ ⟶n=23 \\ </p><p></p><p>\: \: \: \: \: \: \rm\longrightarrow \fbox \pink{n = 1.5} { \boxed{\mathbb{\orange{⛄answer}}}}\\ ⟶n = 1.5</p><p></p><p>$

Explanation:

$hope \: it \: help \: fulfor \: you$