Understanding the Thin Lens Formula in Physics

The thin lens formula provides the mathematical relationship among the object distance, image distance, and focal length for a thin lens under the paraxial approximation, which is fundamental for geometrical optics involving both converging and diverging lenses.

Mathematical Statement of the Thin Lens Formula

For a lens whose thickness is negligible compared to the object and image distances from the lens, the thin lens formula is given by the following equation: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where $f$ denotes the focal length of the lens, $v$ denotes the image distance measured from the optical center of the lens, and $u$ denotes the object distance measured from the same point. The sign convention used is the Cartesian sign convention, consistent with NCERT and JEE standards.

Detailed Derivation of the Thin Lens Formula for a Convex Lens

Consider a thin convex lens placed in air. Let the optical center of the lens be $O$. A real object AB is placed perpendicular to the principal axis at a distance $u$ from $O$.

Let the two spherical surfaces of the lens have radii of curvature $R_1$ (first surface, facing the object) and $R_2$ (second surface, facing away from the object). Let the refractive index of the lens material be $\mu_2$ and that of the surrounding medium (air) be $\mu_1 = 1$.

First, apply the formula for refraction at a spherical surface, which relates object distance, image distance, and radius of curvature at a refracting spherical interface: \[ \frac{\mu_2}{v'} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1} \] where $u$ is the object distance and $v'$ is the image distance formed by the first surface.

For the first surface, the object is at a distance $u$ (from $O$), and the image is formed at $v'$ inside the lens. Substitute $\mu_1 = 1$, $\mu_2 = \mu$, $R_1$: \[ \frac{\mu}{v'} - \frac{1}{u} = \frac{\mu - 1}{R_1} \]

Now, consider the second surface. The image $v'$ formed by the first surface acts as a virtual object for the second surface, which has radius $R_2$ (sign to be taken as per convention). For this surface, the medium changes from the lens ($\mu_2 = \mu$) into air ($\mu_1 = 1$):

\[ \frac{1}{v} - \frac{\mu}{v'} = \frac{1 - \mu}{R_2} \]

To eliminate $v'$ (the intermediate image position inside the lens), solve the previous two equations stepwise. Start with the first equation: \[ \frac{\mu}{v'} = \frac{1}{u} + \frac{\mu - 1}{R_1} \] So, \[ \frac{\mu}{v'} = \frac{1}{u} + \frac{\mu - 1}{R_1} \] \[ \Rightarrow \frac{1}{v'} = \frac{1}{\mu}\left(\frac{1}{u} + \frac{\mu - 1}{R_1}\right) \]

Now substitute $\frac{1}{v'}$ into the second equation: \[ \frac{1}{v} - \frac{\mu}{v'} = \frac{1 - \mu}{R_2} \] Multiply both sides by $1$ (no change), and rearrange: \[ \frac{1}{v} = \frac{\mu}{v'} + \frac{1 - \mu}{R_2} \]

Insert the value of $\frac{\mu}{v'}$ from the previous result: \[ \frac{1}{v} = \left[ \frac{1}{u} + \frac{\mu - 1}{R_1} \right] + \frac{1 - \mu}{R_2} \]

Open the brackets: \[ \frac{1}{v} = \frac{1}{u} + \frac{\mu - 1}{R_1} + \frac{1 - \mu}{R_2} \] Combine $\frac{\mu - 1}{R_1}$ and $\frac{1 - \mu}{R_2}$: \[ \frac{1}{v} = \frac{1}{u} + (\mu - 1)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \]

Bring $\frac{1}{u}$ to the left side: \[ \frac{1}{v} - \frac{1}{u} = (\mu - 1)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \]

Define the focal length $f$ of the thin lens such that when the object is at infinity ($u \to -\infty$), the image is formed at $v = f$. Substitute $u = -\infty$: \[ \frac{1}{f} - 0 = (\mu - 1)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \] Thus, \[ \frac{1}{f} = (\mu - 1)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \]

Now, substitute this into the previously obtained relation: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Rearrange to obtain the standard thin lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] This relation holds for both concave and convex lenses, provided the correct Cartesian sign conventions are used for $u$, $v$, and $f$.

Application of the Thin Lens Formula for Concave (Diverging) and Convex (Converging) Lenses

For a convex (converging) lens, the focal length $f$ is positive as per the sign convention. For a concave (diverging) lens, the focal length $f$ is negative. The same formula is applied, but one must assign the respective signs to the distances and focal length in each case, consistent with the Cartesian convention.

When the object is placed on the left of the lens (real object), $u$ is negative. If the image forms on the right of the lens (real image), $v$ is positive. For a virtual image on the same side as the object, $v$ is negative. This convention is essential when applying the thin lens formula in numerical calculations and theoretical analysis.

Focal Power and the Thin Lens Formula for Lens Strength

The power of a lens, denoted by $P$, quantifies the strength of the lens to converge or diverge rays. It is given by the reciprocal of the focal length in meters: \[ P = \frac{1}{f \ (\text{in meters})} \] The unit of lens power is diopter (D), where $1\, \text{D} = 1\, \text{m}^{-1}$. Convex lenses have positive power, concave lenses have negative power.

Steps in Solving Problems Using the Thin Lens Formula

To use the thin lens formula in problem solving:

1. Assign the correct signs to $u$, $v$, and $f$ as per the sign convention. 2. Substitute the known values into $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. 3. Rearrange algebraically to solve for the unknown quantity. 4. For power, apply $P = 1/f$ with $f$ in meters.

Worked Example: Image Location Using the Thin Lens Formula

Given: An object is placed $15\, \text{cm}$ left of a converging lens of focal length $10\, \text{cm}$.

Substitution: Here, $u = -15\,\text{cm}$ (object is on the left), $f = +10\,\text{cm}$, $v =$ To be found. The formula is: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-15} \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \]

Simplification: \[ \frac{1}{v} = \frac{1}{10} - \frac{1}{15} \] Find a common denominator (30): \[ \frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \]

Final Result: $v = +30\, \text{cm}$. The image is formed $30\,\text{cm}$ on the right side of the lens (real image).

Worked Example: Application for Concave (Diverging) Lens

Given: An object lies $20\, \text{cm}$ to the left of a concave lens of focal length $15\, \text{cm}$ (diverging lens).

Substitution: For a concave lens, $f = -15\,\text{cm}$, $u = -20\,\text{cm}$, $v =$ To be found. The formula is: \[ \frac{1}{-15} = \frac{1}{v} - \frac{1}{-20} \] \[ -\frac{1}{15} = \frac{1}{v} + \frac{1}{20} \]

Simplification: \[ \frac{1}{v} = -\frac{1}{15} - \frac{1}{20} \] \[ \frac{1}{v} = -\left(\frac{4}{60} + \frac{3}{60}\right) \] \[ \frac{1}{v} = -\frac{7}{60} \]

Final Result: $v = -\frac{60}{7} \approx -8.57\, \text{cm}$. The image forms $8.57\,\text{cm}$ to the left of the lens (virtual image).