
A thin symmetrical double convex lens of refractive index \[{\mu _2} = \]1.5 is placed between a medium of refractive index ${\mu _1} = $1.4 to the left and another medium of refractive index ${\mu _3} = $1.6 to the right. Then, the system behaves as.
A. A convex lens
B. A concave lens
C. A glass plate
D. A convo concave lens
Answer
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Hint: We can consider the refraction through the lens as a refraction through a spherical surface twice i.e. once from the front side of the lens and then from the second circular surface of the lens. Refraction through a circular surface is given as$ - \dfrac{{{\mu _1}}}{u} + \dfrac{{{\mu _1}}}{v} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$.
If the rays coming from infinity meet at infinity the lens acts as a glass plate.
Complete step by step answer:
Let us consider two cases for refraction through the two surfaces of the lens. We know that the equation for refraction through a spherical surface is given by as-$ - \dfrac{{{\mu _1}}}{u} + \dfrac{{{\mu _1}}}{v} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$ -----------(1)
Case 1: When the rays travel from first medium$\left( {{\mu _1} = 1.4} \right)$ to the first surface of the lens$\left( {{\mu _2} = 1.5} \right)$.
Substituting the values in equation 1,
$
- \dfrac{{1.4}}{{ - \infty }} + \dfrac{{1.5}}{v} = \dfrac{{1.5 - 1.4}}{R} \\
\Rightarrow 0 + \dfrac{{1.5}}{v} = \dfrac{{0.1}}{R} \\
\Rightarrow v = 15R \\
$
Case 2: When the rays travel from the second surface of the lens$\left( {{\mu _1} = 1.5} \right)$ to the second medium$\left( {{\mu _2} = 1.6} \right)$. The image distance found in case 1 is used as object distance for this case, u$ = 15R$.
Substituting the values in equation 1.
$
- \dfrac{{1.5}}{{15R}} + \dfrac{{1.6}}{V} = \dfrac{{1.6 - 1.5}}{{ - R}} \\
\Rightarrow - \dfrac{{0.1}}{R} + \dfrac{{1.6}}{V} = - \dfrac{{0.1}}{R} \\
\Rightarrow \dfrac{{1.6}}{V} = 0 \\
\Rightarrow \dfrac{1}{V} = 0 \Rightarrow V = \infty \\
$
Therefore the final image distance is $V = \infty $.
Hence the correct option is C the system acts as a glass plate.
Note: Refraction through a lens with two different mediums on its either side can be considered as refraction from a spherical surface twice. The sign of object distance, image distance and focal length should be taken carefully according to the convention to avoid error in answer.
If the rays coming from infinity meet at infinity the lens acts as a glass plate.
Complete step by step answer:
Let us consider two cases for refraction through the two surfaces of the lens. We know that the equation for refraction through a spherical surface is given by as-$ - \dfrac{{{\mu _1}}}{u} + \dfrac{{{\mu _1}}}{v} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$ -----------(1)
Case 1: When the rays travel from first medium$\left( {{\mu _1} = 1.4} \right)$ to the first surface of the lens$\left( {{\mu _2} = 1.5} \right)$.
Substituting the values in equation 1,
$
- \dfrac{{1.4}}{{ - \infty }} + \dfrac{{1.5}}{v} = \dfrac{{1.5 - 1.4}}{R} \\
\Rightarrow 0 + \dfrac{{1.5}}{v} = \dfrac{{0.1}}{R} \\
\Rightarrow v = 15R \\
$
Case 2: When the rays travel from the second surface of the lens$\left( {{\mu _1} = 1.5} \right)$ to the second medium$\left( {{\mu _2} = 1.6} \right)$. The image distance found in case 1 is used as object distance for this case, u$ = 15R$.
Substituting the values in equation 1.
$
- \dfrac{{1.5}}{{15R}} + \dfrac{{1.6}}{V} = \dfrac{{1.6 - 1.5}}{{ - R}} \\
\Rightarrow - \dfrac{{0.1}}{R} + \dfrac{{1.6}}{V} = - \dfrac{{0.1}}{R} \\
\Rightarrow \dfrac{{1.6}}{V} = 0 \\
\Rightarrow \dfrac{1}{V} = 0 \Rightarrow V = \infty \\
$
Therefore the final image distance is $V = \infty $.
Hence the correct option is C the system acts as a glass plate.
Note: Refraction through a lens with two different mediums on its either side can be considered as refraction from a spherical surface twice. The sign of object distance, image distance and focal length should be taken carefully according to the convention to avoid error in answer.
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