- Recap: Lens Equation
- $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
- $f$ = focal length of the lens
- $v$ = image distance
- $u$ = object distance
- Power of a Lens
- The power of a lens is defined as the reciprocal of its focal length.
- Power ($P$) = $\frac{1}{f}$
- Unit: Dioptre (D)
- Example 1: Convex Lens
- Consider a convex lens with a focal length of 10 cm.
- Find the power of the lens.
- $P = \frac{1}{f} = \frac{1}{10} = 0.1$ D
- Example 2: Concave Lens
- Consider a concave lens with a focal length of -12 cm.
- Find the power of the lens.
- $P = \frac{1}{f} = \frac{1}{-12} = -0.083$ D
- Combination of Thin Lenses in Contact
- When two thin lenses are placed in contact with each other, the effective focal length ($f$) and power ($P$) can be calculated using the formula:
- $\frac{1}{f} = \Sigma{\frac{1}{f_i}}$
- $P = \Sigma{P_i}$
- Example 3: Combination of Convex and Concave Lens
- Consider a convex lens with a focal length of 20 cm and a concave lens with a focal length of -15 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{20} + \frac{1}{-15} = -0.033$ cm$^{-1}$
- $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{20} + \frac{1}{-15} = -0.033$ D
- Example 4: Combination of Two Convex Lenses
- Consider two convex lenses with focal lengths of 30 cm and 15 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{15} = 0.05$ cm$^{-1}$
- $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{15} = 0.05$ D
- Example 5: Combination of Two Concave Lenses
- Consider two concave lenses with focal lengths of -25 cm and -10 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{-25} + \frac{1}{-10} = -0.06$ cm$^{-1}$
- $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{-25} + \frac{1}{-10} = -0.06$ D
- Recap: Lens Equation
- $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
- $f$ = focal length of the lens
- $v$ = image distance
- $u$ = object distance
- Power of a Lens
- The power of a lens is defined as the reciprocal of its focal length.
- Power ($P$) = $\frac{1}{f}$
- Unit: Dioptre (D)
- Example 1: Convex Lens
- Consider a convex lens with a focal length of 10 cm.
- Find the power of the lens.
- $P = \frac{1}{f} = \frac{1}{10} = 0.1$ D
- Example 2: Concave Lens
- Consider a concave lens with a focal length of -12 cm.
- Find the power of the lens.
- $P = \frac{1}{f} = \frac{1}{-12} = -0.083$ D
- Combination of Thin Lenses in Contact
- When two thin lenses are placed in contact with each other, the effective focal length and power can be calculated.
- $\frac{1}{f} = \Sigma{\frac{1}{f_i}}$
- $P = \Sigma{P_i}$
- Example 3: Combination of Convex and Concave Lens
- Consider a convex lens with a focal length of 20 cm and a concave lens with a focal length of -15 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{20} + \frac{1}{-15} = -0.033$ cm$^{-1}$
- $P = \frac{1}{f} = -0.033$ D
- Example 4: Combination of Two Convex Lenses
- Consider two convex lenses with focal lengths of 30 cm and 15 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{30} + \frac{1}{15} = 0.05$ cm$^{-1}$
- $P = \frac{1}{f} = 0.05$ D
- Example 5: Combination of Two Concave Lenses
- Consider two concave lenses with focal lengths of -25 cm and -10 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{-25} + \frac{1}{-10} = -0.06$ cm$^{-1}$
- $P = \frac{1}{f} = -0.06$ D
- Recap: Lens Equation
- $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
- $f$ = focal length of the lens
- $v$ = image distance
- $u$ = object distance
- Power of a Lens
- The power of a lens is defined as the reciprocal of its focal length.
- Power ($P$) = $\frac{1}{f}$
- Unit: Dioptre (D)
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- Recap: Lens Equation
- $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
- $f$ = focal length of the lens
- $v$ = image distance
- $u$ = object distance
- Power of a Lens
- The power of a lens is defined as the reciprocal of its focal length.
- Power ($P$) = $\frac{1}{f}$
- Unit: Dioptre (D)
- Example 1: Convex Lens
- Consider a convex lens with a focal length of 10 cm.
- Find the power of the lens.
- $P = \frac{1}{f} = \frac{1}{10} = 0.1$ D
- Example 2: Concave Lens
- Consider a concave lens with a focal length of -12 cm.
- Find the power of the lens.
- $P = \frac{1}{f} = \frac{1}{-12} = -0.083$ D
- Combination of Thin Lenses in Contact
- When two thin lenses are placed in contact with each other, the effective focal length and power can be calculated.
- $\frac{1}{f} = \Sigma{\frac{1}{f_i}}$
- $P = \Sigma{P_i}$
- Example 3: Combination of Convex and Concave Lens
- Consider a convex lens with a focal length of 20 cm and a concave lens with a focal length of -15 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{20} + \frac{1}{-15} = -0.033$ cm$^{-1}$
- $P = \frac{1}{f} = -0.033$ D
- Example 4: Combination of Two Convex Lenses
- Consider two convex lenses with focal lengths of 30 cm and 15 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{30} + \frac{1}{15} = 0.05$ cm$^{-1}$
- $P = \frac{1}{f} = 0.05$ D
- Example 5: Combination of Two Concave Lenses
- Consider two concave lenses with focal lengths of -25 cm and -10 cm, placed in contact with each other.
- Find the effective focal length and power of the combination.
- $\frac{1}{f} = \frac{1}{-25} + \frac{1}{-10} = -0.06$ cm$^{-1}$
- $P = \frac{1}{f} = -0.06$ D
- Recap: Lens Equation
- $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
- $f$ = focal length of the lens
- $v$ = image distance
- $u$ = object distance
- Power of a Lens
- The power of a lens is defined as the reciprocal of its focal length.
- Power ($P$) = $\frac{1}{f}$
- Unit: Dioptre (D)
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