Two concave lenses L-1 and L-2 are kept in contact with each other- If the space between the two lenses is filled with material of refractive index mu - 1- the magnitude of focal length of the combinationBecomes undefinedRemains unchangedIncreasesDecreases

Question

Toppr Admin · Accepted Answer

Let us assume if two identical lens are combined then effective focal length of the lens become half-1fnet-1f1f-2ffnet-f2Hence combination of lens decreases the net focal llength-

Two concave lenses $L_{1}$ and $L_{2}$ are kept in contact with each other. If the space between the two lenses is filled with material of refractive index $μ = 1$ , the magnitude of focal length of the combination
Becomes undefined
Remains unchanged
Increases
Decreases

Let us assume if two identical lens are combined then effective focal length of the lens become half.
$\frac{1}{f_{n e t}} = {\frac{1}{f}}_{\frac{1}{f}} = \frac{2}{f}$
$f_{n e t} = \frac{f}{2}$
Hence combination of lens decreases the net focal llength.

Two concave lenses L1 and L2 are kept in contact with each other. If the space between the two lenses is filled with material of refractive index μ=1, the magnitude of focal length of the combinationBecomes undefinedRemains unchangedIncreasesDecreases

Let us assume if two identical lens are combined then effective focal length of the lens become half.1fnet=1f1f=2ffnet=f2Hence combination of lens decreases the net focal llength.

Two concave lenses $L_{1}$ and $L_{2}$ are kept in contact with each other. If the space between the two lenses is filled with material of refractive index $μ = 1$ , the magnitude of focal length of the combination
Becomes undefined
Remains unchanged
Increases
Decreases

Let us assume if two identical lens are combined then effective focal length of the lens become half.
$\frac{1}{f_{n e t}} = {\frac{1}{f}}_{\frac{1}{f}} = \frac{2}{f}$
$f_{n e t} = \frac{f}{2}$
Hence combination of lens decreases the net focal llength.