A thin converging glass lens made of glass with refractive index 1.5 has a power of $+5.0$ D. When this lens is immersed in a liquid of refractive index $n$, it acts as a divergent lens of focal length 100 cm. What must be the value of $n$?

A thin converging glass lens made of glass with refractive index 1.5 has a power of +5.0 D. When this lens is

Numerical Problems · Q5

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Concept understanding — Lens Maker's Formula

The Intuition: Why a Lens Bends Light

A lens works because light slows down when it enters glass. When a wavefront hits a curved surface at an angle, different parts of it slow down at different moments, and the wavefront bends. The stronger the curvature, the more it bends.

A lens has two surfaces. Each surface bends light by an amount that depends on its radius of curvature $R$ and the refractive index $n$ of the glass. The net bending — the focal length $f$ — is the combined effect of both surfaces.

If you had a single spherical surface separating air from glass, its contribution to bending power is $\frac{n-1}{R}$ . A lens has two such surfaces: light goes from air into glass at the first surface, then from glass back into air at the second. Because the two surfaces face opposite directions relative to the travelling light, their radii typically carry opposite signs.

Note

This uses the New Cartesian Sign Convention (the one used in NCERT and CBSE): all distances are measured from the optical centre, and the direction the incident light travels in is taken as positive. So $R$ is positive if the centre of curvature lies on the side the light is travelling towards (the outgoing side), and negative if it lies on the side the light is travelling from (the incident side).

The Precise Statement

For a thin lens (thickness negligible compared to the radii), the Lens Maker's Formula is:

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

where:

$f$ is the focal length of the lens (positive for converging, negative for diverging)
$n$ is the refractive index of the lens material relative to the surrounding medium (usually air)
$R_1$ is the radius of curvature of the first surface (the one light reaches first)
$R_2$ is the radius of curvature of the second surface

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

How to Apply It: A Worked Example

Take a biconvex lens made of glass ( $n = 1.5$ ) with both surfaces having the same radius of curvature magnitude, $20$ cm.

Light travels left to right. The first surface bulges toward the incoming light, so its centre of curvature lies to the right of the surface — on the side the light is travelling towards. By the rule above, $R_1 = +20$ cm.

The second surface also bulges outward (away from the lens), so its centre of curvature lies to the left of that surface — on the side the light is travelling from. So $R_2 = -20$ cm.

Plug in:

\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5 \times \left(\frac{1}{20} + \frac{1}{20}\right) = 0.5 \times \frac{2}{20} = \frac{1}{20}

So $f = +20$ cm. Positive means converging — correct for a biconvex lens.

Watch out

The most common mistake is getting the sign of $R_2$ wrong. For a biconvex lens, $R_1$ is positive and $R_2$ is negative. For a biconcave lens, it's the reverse: $R_1$ negative, $R_2$ positive. Always sketch the lens and mark where each surface's centre of curvature actually sits.

Why the Formula Works (Brief Derivation) …

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Step 1. In air, the lens has power $P_{\text{air}}=+5.0$ D, so $\dfrac{1}{f_{\text{air}}}=(n_g-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)=5.0$ , with $n_g=1.5$ ; this gives the geometric factor $\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)=\dfrac{5.0}{1.5-1}=\dfrac{5.0}{0.5}=10\ \text{m}^{-1}$ .

Step 2. In the liquid of unknown index $n$ , the same lens has focal length $-100$ cm $=-1$ m, i.e. $P_{\text{liquid}}=-1$ D, so $\dfrac{1}{f_{\text{liquid}}}=\left(\dfrac{n_g}{n}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)=-1$ .

Step 3. Substituting the geometric factor found in Step 1: $\left(\dfrac{1.5}{n}-1\right)\times10=-1$ , so $\dfrac{1.5}{n}-1=-0.1$ , i.e. $\dfrac{1.5}{n}=0.9$ . …

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Q.A thin converging glass lens made of glass with refractive index 1.5 has a power of +5.0+5.0+5.0 D. When this lens is immersed in a liquid of refractive index nnn, it acts as a divergent lens of focal length 100 cm. What must be the value of nnn?

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Q.A thin converging glass lens made of glass with refractive index 1.5 has a power of $+5.0$ D. When this lens is immersed in a liquid of refractive index $n$ , it acts as a divergent lens of focal length 100 cm. What must be the value of $n$ ?

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