The graph between $$\dfrac{1}{u}$$ and $$\dfrac{1}{v}$$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is $$1.5$$ and its both the surfaces have same radius of curvatures $$R$$. The value of $$R$$ will be ______ cm. (Where $$u$$ = object distance, $$v$$ = image distance)

We need to find the value of the radius of curvature $$R$$ for a thin equi-convex lens based on the given graph between $$\frac{1}{v}$$ and $$\frac{1}{u}$$.

1. Relate the Graph to the Lens Formula

The standard lens formula for a thin lens relates the object distance ($$u$$), image distance ($$v$$), and focal length ($$f$$):

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

Rearranging this equation to express it in the slope-intercept straight-line form ($$y = mx + c$$), where $$\frac{1}{v}$$ is plotted on the vertical $$y$$-axis and $$\frac{1}{u}$$ is plotted on the horizontal $$x$$-axis:

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

From this equation:

• When the line intersects the vertical axis ($$\frac{1}{u} = 0$$ at point $$A$$), the $$y$$-intercept is equal to $$\frac{1}{f}$$.
• When the line intersects the horizontal axis ($$\frac{1}{v} = 0$$ at point $$B$$), the $$x$$-intercept is equal to $$-\frac{1}{f}$$.

2. Extract the Focal Length ($$f$$) from the Graph

Looking at the graph provided :

• The vertical intercept at point $$A$$ has a value of:

  $$\frac{1}{v} = 0.10\text{ cm}^{-1}$$

Equating this intercept to our formula gives:

$$\frac{1}{f} = 0.10\text{ cm}^{-1} \implies f = \frac{1}{0.10} = 10\text{ cm}$$

3. Calculate the Radius of Curvature ($$R$$)

According to the Lens Maker's Formula for a thin lens:

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

For an equi-convex lens under standard Cartesian sign conventions:

• The radius of curvature of the first surface is positive ($$R_1 = +R$$).
• The radius of curvature of the second surface is negative ($$R_2 = -R$$).

Substituting these along with the given refractive index ($$n = 1.5$$):

$$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right)$$

$$\frac{1}{f} = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R}$$

Since $$\frac{1}{f} = \frac{1}{R}$$, it directly simplifies to:

$$R = f = 10\text{ cm}$$

Conclusion

The value of the radius of curvature $$R$$ is 10 cm.