Q.An equiconvex lens of focal length f is split in such a way that two plano-convex lenses are formed. Find the focal length of each.

Showing the 12 most recent of 22 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.A concave lens of focal length $10\ \text{cm}$ is cut into two identical plano-concave lenses. The focal length of each lens will be (A) $20\ \text{cm}$ (B) $30\ \text{cm}$ (C) $40\ \text{cm}$ (D) $5\ \text{cm}$

   ›Reveal solutionSolution

   Cutting a symmetric concave lens through its middle (perpendicular to the principal axis) leaves each piece with only one curved surface, so each plano-concave piece has half the power — and therefore double the focal length: 20 cm, option (A).

   Concept and Intuition

   The lens maker's formula relates a thin lens's focal length to its two radii of curvature and the refractive index of the material:

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

   A symmetric biconcave lens has two curved surfaces, and each contributes equally to the total power. When the lens is cut through its middle by a plane perpendicular to the principal axis, each piece keeps one original curved surface and gains a flat face. A flat surface has an infinite radius of curvature, so it contributes nothing to the power — each piece is left with only half the original bending power.

   Step-by-Step Solution

   1. Apply the formula to the original biconcave lens.

   Use the New Cartesian sign convention with light travelling left to right. For a biconcave lens of equal radii of magnitude $R$: the first surface's centre of curvature lies on the incident (left) side, so $R_1 = -R$; the second surface's centre of curvature lies on the outgoing (right) side, so $R_2 = +R$. Then

   $\frac{1}{f} = (\mu-1)\left(\frac{1}{-R} - \frac{1}{+R}\right) = -\frac{2(\mu-1)}{R}$

   The negative sign confirms a diverging lens. With $f = -10\ \text{cm}$:

   $\frac{\mu-1}{R} = \frac{1}{20}\ \text{cm}^{-1}$

   2. Apply it to one plano-concave piece.

   Each piece keeps one curved surface ($R_1 = -R$) and gains a flat cut face ($R_2 = \infty$):

   $\frac{1}{f'} = (\mu-1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) = -\frac{\mu-1}{R} = -\frac{1}{20}\ \text{cm}^{-1}$

   3. Read off the result.

   $f' = -20\ \text{cm}$ …

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2. CBSE 2026Set 55/3/11 markMCQ
   Q.A thin plano-convex lens and a thin equi-concave lens are kept coaxially in contact as shown in the figure. Assuming both the lenses are made of glass of refractive index $\mu$, and $R$ is the radius of curvature of each curved surface, the focal length of the combination is : (A) $\dfrac{R}{\mu-1}$ (B) $-\dfrac{R}{\mu-1}$ (C) $\dfrac{2R}{\mu-1}$ (D) $-\dfrac{2R}{\mu-1}$

   ›Reveal solutionSolution

   We calculate the focal lengths of the plano-convex and equi-concave lenses separately using the lens maker's formula, applying the correct sign conventions for radii of curvature. Then, we combine these focal lengths to find the equivalent focal length of the system. The focal length of the combination is $\boxed{-\dfrac{R}{\mu-1}}$.

   

   Concept and Intuition

   To find the focal length of a combination of thin lenses kept in contact, we first need to determine the focal length of each individual lens. The fundamental tool for this is the Lens Maker's Formula.

   Lens Maker's Formula

   The focal length $f$ of a thin lens made of a material with refractive index $\mu$ (relative to the surrounding medium, usually air, for which $\mu_{air} = 1$) is given by:

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

   Here, $R_1$ is the radius of curvature of the first surface encountered by light, and $R_2$ is the radius of curvature of the second surface. The signs of $R_1$ and $R_2$ are crucial and follow a specific convention.

   Sign Convention for Radii of Curvature

   We will use the following convention for $R_1$ and $R_2$ in the lens maker's formula, assuming light travels from left to right:

   • $R_1$ (First Surface):
     • If the first surface is convex (bulges towards the right), $R_1$ is positive ($+R$).
     • If the first surface is concave (bulges towards the left), $R_1$ is negative ($-R$).
     • If the first surface is flat (plano), $R_1$ is infinite ($\infty$).
   • $R_2$ (Second Surface):
     • If the second surface is convex (bulges towards the left), $R_2$ is negative ($-R$).
     • If the second surface is concave (bulges towards the right), $R_2$ is positive ($+R$).
     • If the second surface is flat (plano), $R_2$ is infinite ($\infty$).

   Watch out

   The sign convention for $R_1$ and $R_2$ is a common source of error. Always be consistent with the convention you choose. The one outlined above ensures that converging lenses have positive focal lengths and diverging lenses have negative focal lengths when $\mu > 1$.

   Combination of Thin Lenses in Contact

   When two thin lenses with focal lengths $f_1$ and $f_2$ are placed coaxially in contact, the focal length $F$ of the combination is given by:

   $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$

   Step-by-step Solution

   1. Identify the properties of the plano-convex lens (Lens 1).

      • Refractive index: $\mu$
      • First surface: Flat. According to our sign convention, $R_1 = \infty$.
      • Second surface: Convex. It bulges towards the left (as seen from the second surface, or its center of curvature is to the left). According to our sign convention, $R_2 = -R$.

   2. Calculate the focal length of the plano-convex lens ($f_1$).

      Using the lens maker's formula:

   $\frac{1}{f_1} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

   Substitute the values for $R_1$ and $R_2$:
   

   $\frac{1}{f_1} = (\mu - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right)$

   $\frac{1}{f_1} = (\mu - 1) \left( 0 + \frac{1}{R} \right)$

   $\frac{1}{f_1} = \frac{\mu - 1}{R}$

   Therefore, the focal length of the plano-convex lens is:
   

   $f_1 = \frac{R}{\mu - 1}$

   This is a positive focal length, as expected for a converging lens.
   

   3. Identify the properties of the equi-concave lens (Lens 2).

   * Refractive index: $\mu$ …

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3. CBSE 2026Set DS1 mark
   Q.In which the power of a lens will be large — in air or water?

   ›Reveal solutionSolution

   Power is larger in air, because the glass–water relative refractive index is smaller than the glass–air one.

   Concept. By the lens-maker's formula the power of a lens depends on the refractive index of the lens relative to its surroundings:

   $P = \frac1f = \left({}_{m}n_g - 1\right)\left(\frac1{R_1}-\frac1{R_2}\right),$

   where ${}_{m}n_g = n_g/n_m$ is the index of glass with respect to the medium.

   • In air ($n_m \approx 1$): ${}_{a}n_g \approx 1.5$, so $({}_{a}n_g-1) \approx 0.5$.
   • In water ($n_m \approx 1.33$): ${}_{w}n_g = 1.5/1.33 \approx 1.13$, so $({}_{w}n_g-1) \approx 0.13$. …

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4. CBSE 2025Set D1 markMCQ
   Q.A convex lens is dipped in a liquid, whose refractive index is equal to the refractive index of the material of the lens. Then its focal length will (A) become zero (B) become infinity (C) reduce (D) increase

   ›Reveal solutionSolution

   Lensmaker's formula has a factor (n_lens/n_medium − 1); if the two indices are equal this factor is zero, so 1/f = 0 and f → ∞.

   By the lensmaker's formula in a medium,

   1/f = (n_lens/n_medium − 1)(1/R₁ − 1/R₂)

   If the liquid's refractive index equals the lens material's index, then n_lens/n_medium = 1, so the factor (1 − 1) = 0.

   …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.When monochromatic red light is used instead of blue light in a convex lens, its focal length

   (a) does not change

   (b) increases

   (c) decreases

   (d) remain same

   ›Reveal solutionSolution

   Red light has a lower refractive index than blue (dispersion), and $f \propto 1/(n-1)$, so lower $n$ gives a larger $f$.

   By the lens maker's formula, $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, so $f \propto \dfrac{1}{(n-1)}$ for fixed geometry. Due to dispersion, the refractive index of a material is slightly higher for blue light than for red light ($n_{blue} > n_{red}$, since blue light bends more). Using red …

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6. CBSE 2024Set FS1 mark
   Q.Find the ratio of focal length of lens in air and that of lens when it is immersed in liquid.

   ›Reveal solutionSolution

   $\dfrac{f_{air}}{f_{liquid}}=\dfrac{n_g-n_l}{n_l\,(n_g-1)}$, where $n_g$, $n_l$ are the refractive indices of glass and the liquid.

   Concept. The lens maker's formula uses the index of the lens relative to its surroundings:

   $\frac{1}{f}=\left(\,{}_{med}n_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$

   In air ($n_{air}=1$): $\dfrac{1}{f_{air}}=(n_g-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right).$

   In liquid: the glass index relative to the liquid is $n_g/n_l$, so …

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7. CBSE 2024Set ANNUAL1 mark
   Q.The power of a lens is greater in water or air?

   ›Reveal solutionSolution

   A glass lens is more powerful in air than in water, because water's refractive index is closer to that of glass.

   Power of a lens, $P = 1/f$, and from the lens maker's formula:

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

   where ${}_a\mu_m$ is the refractive index of the surrounding medium (relative to air).

   • In air, ${}_a\mu_m = 1$, so the relative refractive index of glass w.r.t. the medium is large ($\approx 1.5$), giving a small $f$ and large $P$. …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.The refractive index of the material of a double equiconvex lens is 2.5. If R be its radius of curvature, then its focal length is

   (a) 0

   (b) R/3

   (c) 2R

   (d) 3R.

   ›Reveal solutionSolution

   Applying the lens maker's formula to an equiconvex lens (equal radii of curvature, opposite sign) with refractive index 2.5 gives $f = R/3$.

   The lens maker's formula is

   $\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

   For a double equiconvex lens, both surfaces bulge outward with the same radius of curvature magnitude $R$. Using the standard sign convention (distances measured from the optical centre, in the direction of incident light being positive): the first surface is convex towards the incoming light, so $R_1 = +R$; the second surface's centre of …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The focal length of a glass (μ=1.5) lens in air is 20cm. If it is dipped in water (μ = 4/3), its focal length in water will be -

   (a) 80 cm

   (b) 40 cm

   (c) 60 cm

   (d) 20 cm

   ›Reveal solutionSolution

   A lens's focal length depends on its refractive index relative to the surrounding medium; a converging lens weakens (longer $f$) when moved from air into a denser medium like water.

   Lensmaker's equation: $\dfrac{1}{f} = \left(\dfrac{\mu_{lens}}{\mu_{medium}} - 1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$

   In air: $\mu_{rel} = 1.5/1 = 1.5$

   $\dfrac{1}{20} = (1.5-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) \Rightarrow \left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) = \dfrac{1}{10}$

   …

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10. CBSE 2023Set BS1 mark
    Q.What will be the effect on the focal length and nature of a convex lens of glass of refractive index $n = \dfrac{3}{2}$ dipped in a liquid of refractive index $n = \dfrac{3}{2}$?

    ›Reveal solutionSolution

    With glass and surrounding liquid of the same index, $f\to\infty$; the lens stops acting as a lens.

    The lens-maker's formula in a medium uses the relative refractive index $n_{rel}=\dfrac{n_{glass}}{n_{liquid}}$:

    $\frac{1}{f}=(n_{rel}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$ …

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11. CBSE 2023Set F1 markMCQ
    Q.The radius of curvature of each surface of a biconvex lens is 20 cm and the refractive index of the material of the lens is 1.5. The focal length of the lens is (A) 20 m (B) 1/20 m (C) 20 cm (D) 1/20 cm

    ›Reveal solutionSolution

    Lens maker's formula gives f = 20 cm for a biconvex lens with R = 20 cm and n = 1.5.

    For a thin lens, the lens maker's formula is

    $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

    For a biconvex lens with each surface radius 20 cm, using sign convention $R_1 = +20$ cm and $R_2 = -20$ cm:

    …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.A convex lens is immersed in a liquid of refractive index greater than that of glass. It will behave as a -

    (a) convergent lens

    (b) divergent lens

    (c) plane glass

    (d) homogeneous liquid

    ›Reveal solutionSolution

    When the surrounding medium is optically denser than the lens material, a convex lens's effect reverses.

    …

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