Q.Calculate the radius of curvature of an equi-concave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of $-5\,\text{D}$?

Showing the 12 most recent of 58 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 2026Set A1 markMCQ
   Q.The ratio of the refractive index of red light to blue light in air is (A) less than unity (B) greater than unity (C) equal to unity (D) none of these

   ›Reveal solutionSolution

   Because refractive index rises from red to blue (dispersion), n_red < n_blue, so their ratio is less than 1.

   Dispersion arises because the refractive index of a medium depends on wavelength: it is larger for shorter wavelengths (blue) and smaller for longer wavelengths (red). Thus $n_{red}<n_{blue}$, and their ratio is

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.The focal length of a lens is minimum for which colour?

   (a) Red

   (b) Yellow

   (c) Violet

   (d) Blue

   ›Reveal solutionSolution

   Since refractive index decreases from violet to red (dispersion), and $f\propto 1/(\mu-1)$, the colour with the highest $\mu$ — violet — bends most and has the smallest focal length.

   For a thin lens, the lens-maker's formula gives $\dfrac{1}{f}=(\mu-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, so $f\propto\dfrac{1}{\mu-1}$: a larger refractive index means a smaller focal length. Because the refractive index of ordinary glass is greater for shorter wavelengths, the order from lowest to highest $\mu$ across the visible spectrum is red < orange < yellow < green < blue < …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.The speed of light in an isotropic medium depends on :

   (a) the nature of propagation

   (b) its intensity

   (c) the motion of the source with respect to medium

   (d) its wavelength

   ›Reveal solutionSolution

   A medium's refractive index (and hence the speed of light within it) varies with wavelength -- the phenomenon of dispersion.

   Working

   The speed of light in a medium is $v=c/n$. For any real (non-vacuum) isotropic medium, the refractive index $n$ is not fixed but depends on the wavelength of the light (e.g., violet light travels slightly slower than red light in glass) -- this dependence is the cause of dispersion (splitting of white light by a prism).

   …

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7. 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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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.In glass, the velocity of the light is minimum for :

   (a) red

   (b) violet

   (c) yellow

   (d) green

   ›Reveal solutionSolution

   Violet light bends the most and travels slowest in glass because glass has its highest refractive index for violet light.

   Speed of light in a medium is $v = c/n$. Due to dispersion, the refractive index of glass is different for different colours and increases from red to violet (violet has the shortest wavelength and is refracted most). Since $n_{violet}$ is the largest …

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9. 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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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.If red light and violet rays are of focal lengths f_r and f_v, then which one of the following is true ?

    (a) lambda_r <= lambda_v

    (b) lambda_r = lambda_v

    (c) mu_r > mu_v

    (d) mu_r < mu_v

    ›Reveal solutionSolution

    Violet light has a shorter wavelength and is refracted more, so its refractive index is higher than red light's.

    By Cauchy's dispersion relation, the refractive index of a medium decreases as wavelength increases: $\mu = A + \dfrac{B}{\lambda^2}$ (approximately). Since red light has a longer wavelength than violet light ($\lambda_r > \lambda_v$), red light is refracted less and has a smaller refractive index than violet light.

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.If the refractive index of a material of equilateral prism is sqrt(3), then the angle of minimum deviation is

    (i) 30 degrees

    (ii) 45 degrees

    (iii) 60 degrees

    (iv) 75 degrees

    ›Reveal solutionSolution

    Dm = 60 degrees for an equilateral prism with mu = sqrt(3).

    The prism formula is $\mu = \frac{\sin\!\left(\frac{A + D_m}{2}\right)}{\sin\!\left(\frac{A}{2}\right)}$. For an equilateral prism $A = 60^\circ$, so $\sin(A/2) = \sin 30^\circ = 0.5$. Then …

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12. CBSE 2024Set FS1 markMCQ
    Q.The maximum focal length of convex lens is for:

    (i) blue light

    (ii) green light

    (iii) red light

    (iv) yellow light

    ›Reveal solutionSolution

    $\dfrac{1}{f}\propto(n-1)$; red light has the least refractive index in glass, so it gives the greatest focal length — option (iii).

    Concept. For a lens, $\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, so $f$ is largest when the refractive index $n$ is smallest.

    …

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