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SMA Physics โ€ข Physics Lab Class XII โ€ข Exp 18

๐Ÿ” Focal Length of Convex Lens (uโ€“v Graphs)

๐Ÿ“– Manual
๐Ÿ“ Optical Bench: Ray Refraction & Parallax Alignment
๐Ÿ’ก
Experiment Guide: Select an Object Distance ($u$). Move the Image Needle ($v_{\text{needle}}$) on the other side of the lens until it meets the tip of the inverted real image. Slide the Observer Eye to confirm zero parallax!
Object Distance $u$:30.0 cm
Theoretical Image $v$:30.0 cm
Image Needle Pos:30.0 cm
Parallax Status:ZERO PARALLAX โœ“
๐ŸŽ›๏ธ Optical Bench Uprights Lens Focal Length $f = 15.0\text{ cm}$
Object Needle Distance ($u$) 30.0 cm
Range: 18.0 cm to 60.0 cm ($u > f$) Real inverted image
Image Needle Upright ($v_{\text{needle}}$) 30.0 cm
Position on opposite side of lens Coincide with image tip
Observer Eye Transverse Shift (Parallax Test) Center
โ† Left Tilt Direct Center Right Tilt โ†’
๐Ÿ“ Computed Lens Parameters
Focal Length $f = \frac{uv}{u+v}$
15.0 cm
Lens Power $P = 100/f$
+6.67 D
Magnification $m = -v/u$
-1.00
Image Characteristics
Real & Inverted

Observation Table: Convex Lens Focal Length

Formula: $f = \frac{uv}{u+v}$ (using absolute magnitudes). Power $P = 100/f$ diopters.

S.No. $u$ (cm) $v$ (cm) $1/u$ (cmโปยน) $1/v$ (cmโปยน) Focal Length $f = \frac{uv}{u+v}$ (cm) Parallax Check
๐Ÿ“Š Experimental Mean Result
Mean Focal Length ($f$)
-- cm
Mean Lens Power ($P$)
-- D

Optics Graphs: $u - v$ Curve & $1/u - 1/v$ Straight Line

Line $u=v$ intersects hyperbola at $(2f, 2f) = (30, 30)$. Straight line has intercepts $= 1/f$.

โ„น๏ธ
In the $u-v$ plot, draw a line from the origin at $45^\circ$ ($u = v$). The point of intersection gives $u = v = 2f = 30\text{ cm}$, giving $f = 15.0\text{ cm}$. In the $1/u - 1/v$ straight-line graph, both the $x$ and $y$ intercepts equal $1/f = 0.067\text{ cm}^{-1}$.

๐ŸŽฏ Aim of the Experiment

To find the focal length of a convex lens by plotting graphs between $u$ and $v$ or between $1/u$ and $1/v$.

๐Ÿ“ Lens Formula & Graph Methods

The relation between object distance $u$, image distance $v$, and focal length $f$ of a lens is:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

Applying Cartesian sign conventions (incident light left to right: $u$ is negative, $v$ is positive for real image):

$$\frac{1}{f} = \frac{1}{v} - \left(-\frac{1}{u}\right) = \frac{1}{v} + \frac{1}{u} \implies f = \frac{u \cdot v}{u + v}$$

Graphical Evaluation:

  • $u-v$ Method: Plot $u$ on horizontal axis and $v$ on vertical axis. Draw bisector line $u = v$. The intersection point has coordinates $(2f, 2f)$, hence $f = \text{coordinate} / 2$.
  • $1/u-1/v$ Method: Plot $1/u$ on $x$-axis and $1/v$ on $y$-axis. The line has slope $-1$, and intercepts on both axes equal $OA = OB = 1/f \implies f = 1/OA$.

โ“ Interactive Viva Voce Preparation