1} - \frac{1}{R_2}\right)$$

For the original biconvex lens, both radii of curvature have the same magnitude but opposite signs, so let's denote them as ±R. The focal length of the original lens is given as 10 cm, and the refractive index (n) is constant for both the original lens and the new plano-convex lenses.

For the original lens, the lensmaker's formula becomes:

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

Since the focal length is given as 10 cm, we can plug in the value:

$$\frac{1}{10} = (n - 1) \left(\frac{1}{R} + \frac{1}{R}\right)$$

$$\frac{1}{10} = (n - 1) \left(\frac{2}{R}\right)$$

For each plano-convex lens, one radius of curvature (R1) is infinite (the flat side), and the other radius (R2) is the same as the original lens (R). The lensmaker's formula for the plano-convex lens becomes:

$$\frac{1}{f'} = (n - 1) \left(\frac{1}{\infty} - \frac{1}{R}\right)$$

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

Now, we can substitute the expression for (n-1)(2/R) from the original lens equation:

$$\frac{1}{f'} = \frac{1}{10} \cdot \frac{1}{2}$$ $$\frac{1}{f'} = \frac{1}{20}$$

So, the focal length of each plano-convex lens (f') is 20 cm. To find the power of each lens, we can use the formula:

$$P (\text{in diopters}) = \frac{1}{f (\text{in meters})}$$

Converting the focal length to meters and calculating the power:

$$P = \frac{1}{0.2}$$

$$P = 5 \text{ D}$$

So, the power of each plano-convex lens after the cut is 5 diopters (D).

Geometrical Optics

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