Two identical equiconvex lenses, each of focal length f

Answer

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Hint: We know that when two lenses are kept closer then there will be the concept of combination of lenses and the space between them will act as a lens. Then the focal length of these lenses should be known to solve the problem in a proper manner. Lens makers' formula is a vital need for these types of problems.

Complete step-by-step solution -

Here the condition is such that the combination of two lenses will create one other lens in between them. When they will be kept close to each other and the space between them is filled with water then the little space between them will also act like a lens. These are two convex lenses which are identical (same focal length and same refractive index), when placed close then they will create a concave lens in between them and the refractive index of this newly formed concave lens is given.Given,Refractive index of both lenses= ${\mu _g} = \dfrac{3}{2}$Focal length of both lenses=${f_1} = {f_2} = f$------equation (1)Refractive index of water which is filled in between=${\mu _w} = \dfrac{4}{3}$Now will find the focal length of both lenses separately,Lens maker formula,$\dfrac{1}{f} = (\mu - 1)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$Where ${{\text{R}}_1}$and ${{\text{R}}_2}$are the radius of curvature.For the lens 1,Let the radius of curvature of both the lenses=${\text{R}}$Now applying the lens makers formula for the lens given,Then,$\dfrac{1}{{{f_1}}} = (\dfrac{3}{2} - 1)\left[ {\dfrac{1}{R} - \dfrac{1}{{( - R)}}} \right] \\\ $$ \Rightarrow \dfrac{1}{{{f_1}}} = \dfrac{1}{2}\left[ {\dfrac{1}{R} + \dfrac{1}{R}} \right] \\\Rightarrow \dfrac{1}{{{f_1}}} = \dfrac{1}{R} \\\Rightarrow {f_1} = R \\ \ $So,${f_1} = {f_2} = R$And also it is given that, ${f_1} = {f_2} = f$So we get, $R = f$-----equation (2)Now for the lens which is formed in between them,Let it’s focal length is ${f_3}$,Again by the lens makers formula for this concave lens$\dfrac{1}{{{f_3}}} = (\dfrac{4}{3} - 1)\left[ {\dfrac{1}{{ - R}} - \dfrac{1}{R}} \right] \\\Rightarrow \dfrac{1}{{{f_3}}} = \dfrac{{ - 2}}{{3R}} \\\Rightarrow {f_3} = \dfrac{{ - 3R}}{2} \\ $From equation (1), putting the value of \[R\], we get the following relation,${f_3} = \dfrac{{ - 3f}}{2}$-------equation (3)Now by applying the combination concept of the lenses placed in contact,Let us assume that the equivalent focal length of the combination is ${f_{eq}}$,then following will be the relation,$\dfrac{1}{{{f_{eq}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} + \dfrac{1}{{{f_3}}} + - - - - $On substituting the values from equation (1) and equation (3),We have,$ \Rightarrow $$\dfrac{1}{{{f_{eq}}}} = \dfrac{1}{f} + \dfrac{1}{f} + \dfrac{{ - 2}}{{3f}} + - - - - $$ \Rightarrow $$\dfrac{1}{{{f_{eq}}}} = \dfrac{1}{f} + \dfrac{1}{f} + \dfrac{1}{{\dfrac{{ - 3f}}{2}}}$$ \Rightarrow $$\dfrac{1}{{{f_{eq}}}} = \dfrac{{3 + 3 - 2}}{{3f}}$$ \Rightarrow $$\dfrac{1}{{{f_{eq}}}} = \dfrac{4}{{3f}}$$ \Rightarrow $${f_{eq}} = \dfrac{{3f}}{4}$Hence the focal length of the combination is ${f_{eq}} = \dfrac{{3f}}{4}$,

So the answer to this question is option (D).

Note: Thus by combining the different lenses like this we can generate some lenses and by applying the concept of combination of lenses we can get the equivalent focal length of the combination.The focal length of the convex lens is taken positive while the focal length of the concave lens is taken negative for the calculation.

Two identical glass μ g =3 / 2 equiconvex lenses of focal length f are kept in contact. The space between the two lenses is filled with water μ w =.4 / 3 . The focal length of the combination isA.fB. f2C. 4 f 3D. 3 f 4

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Two thin equi convex lenses each of focal length f and made of glass μ g =3/2 are placed in contact. The space between them is filled with water μ w =4/3. The focal length of the combination is A. 1/2 fB. 2/3C. 4/7D. 3 f /4

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