Two identical thin biconvex lens of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is \underline\hspace2cm cm.

Geometrical Optics PYQ — JEE Main 2022 · 24 Jun · Shift 1 · Q63 with Solution

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Correct answer: 10

Given data

Two identical thin biconvex lenses
Focal length of each lens in air: $f=15\,\text{cm}$
Refractive index of lens material: $\mu_g=1.5$
Refractive index of liquid between lenses: $\mu_l=1.25$

We need the focal length of the combination.

Find the shape factor of each lens

For a thin lens in air,

\frac{1}{f}=(\mu_g-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Since $f=15$ cm and $\mu_g=1.5$ ,

\frac{1}{15}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

\frac{1}{15}=0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{2}{15}

Let

S=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{2}{15}

Power of the outer surfaces

The two-lens system has four refracting surfaces, but the two inner curved surfaces enclose the liquid. Since the lenses are identical and in contact, the middle two surfaces are equal and opposite in curvature. Their powers cancel when the medium on both sides is the same liquid.

So only the two outer surfaces contribute effectively.

For a spherical refracting surface,

P=\frac{n_2-n_1}{R}

First outer surface: air to glass
Last outer surface: glass to air

Thus total power of combination is

P=(\mu_g-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

for one lens-like equivalent using only the outer surfaces.

But here there are two identical lenses, so the outer surfaces together give

P=(\mu_g-1)S=0.5\cdot \frac{2}{15}=\frac{1}{15}

Wait carefully: this corresponds to the contribution of the two outermost surfaces together, which is exactly the same as one lens power, because the inner powers cancel.

Hence

\frac{1}{F}=\frac{1}{15}

would give $F=15$ cm if only cancellation were considered directly. But this misses the fact that each original lens had two surfaces in air, while now the inner surfaces are in liquid and do not fully vanish in system power treatment unless surface-by-surface power is summed correctly.

So let us do it properly by summing all four refracting surface powers.

Power of each surface separately

For a thin system of refracting surfaces in contact, total power is the sum of powers of all surfaces:

P=\sum \frac{n_{\text{after}}-n_{\text{before}}}{R}

Let the radii of one biconvex lens be $R$ and $-R$ .

Since the lenses are identical and facing same way in contact, the four surfaces are:

Surface 1: air to glass, radius $+R$
Surface 2: glass to liquid, radius $-R$
Surface 3: liquid to glass, radius $+R$
Surface 4: glass to air, radius $-R$

Now,

P_1=\frac{1.5-1}{R}=\frac{0.5}{R}

P_2=\frac{1.25-1.5}{-R}=\frac{-0.25}{-R}=\frac{0.25}{R}

P_3=\frac{1.5-1.25}{R}=\frac{0.25}{R}

P_4=\frac{1-1.5}{-R}=\frac{-0.5}{-R}=\frac{0.5}{R}

Therefore total power,

P=\frac{0.5+0.25+0.25+0.5}{R}=\frac{1.5}{R}

Relate $R$ using the focal length of one lens in air

For one original biconvex lens in air,

\frac{1}{15}=(1.5-1)\left(\frac{1}{R}-\frac{1}{-R}\right)

\frac{1}{15}=0.5\left(\frac{2}{R}\right)=\frac{1}{R}

So,

R=15\,\text{cm}

Find focal length of the combination

Now,

P=\frac{1.5}{R}=\frac{1.5}{15}=\frac{1}{10}

Hence,

F=10\,\text{cm}

Final answer

\boxed{10}

The focal length of the combination is $10$ cm.

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View written solutionFree

Correct answer: 10

Given data

Two identical thin biconvex lenses
Focal length of each lens in air: $f=15\,\text{cm}$
Refractive index of lens material: $\mu_g=1.5$
Refractive index of liquid between lenses: $\mu_l=1.25$

We need the focal length of the combination.

Find the shape factor of each lens

For a thin lens in air,

\frac{1}{f}=(\mu_g-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Since $f=15$ cm and $\mu_g=1.5$ ,

\frac{1}{15}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

\frac{1}{15}=0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{2}{15}

Let

S=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{2}{15}

Power of the outer surfaces

So only the two outer surfaces contribute effectively.

For a spherical refracting surface,

P=\frac{n_2-n_1}{R}

First outer surface: air to glass
Last outer surface: glass to air

Thus total power of combination is

P=(\mu_g-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

for one lens-like equivalent using only the outer surfaces.

But here there are two identical lenses, so the outer surfaces together give

P=(\mu_g-1)S=0.5\cdot \frac{2}{15}=\frac{1}{15}

Wait carefully: this corresponds to the contribution of the two outermost surfaces together, which is exactly the same as one lens power, because the inner powers cancel.

Hence

\frac{1}{F}=\frac{1}{15}

So let us do it properly by summing all four refracting surface powers.

Power of each surface separately

For a thin system of refracting surfaces in contact, total power is the sum of powers of all surfaces:

P=\sum \frac{n_{\text{after}}-n_{\text{before}}}{R}

Let the radii of one biconvex lens be $R$ and $-R$ .

Since the lenses are identical and facing same way in contact, the four surfaces are:

Surface 1: air to glass, radius $+R$
Surface 2: glass to liquid, radius $-R$
Surface 3: liquid to glass, radius $+R$
Surface 4: glass to air, radius $-R$

Now,

P_1=\frac{1.5-1}{R}=\frac{0.5}{R}

P_2=\frac{1.25-1.5}{-R}=\frac{-0.25}{-R}=\frac{0.25}{R}

P_3=\frac{1.5-1.25}{R}=\frac{0.25}{R}

P_4=\frac{1-1.5}{-R}=\frac{-0.5}{-R}=\frac{0.5}{R}

Therefore total power,

P=\frac{0.5+0.25+0.25+0.5}{R}=\frac{1.5}{R}

Relate $R$ using the focal length of one lens in air

For one original biconvex lens in air,

\frac{1}{15}=(1.5-1)\left(\frac{1}{R}-\frac{1}{-R}\right)

\frac{1}{15}=0.5\left(\frac{2}{R}\right)=\frac{1}{R}

So,

R=15\,\text{cm}

Find focal length of the combination

Now,

P=\frac{1.5}{R}=\frac{1.5}{15}=\frac{1}{10}

Hence,

F=10\,\text{cm}

Final answer

\boxed{10}

The focal length of the combination is $10$ cm.

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