Change of Focal Length of Lens Placed in Different Medium

When a lens is placed in a medium other than air, its focal length changes. This phenomenon is due to the alteration in the refractive index of the surrounding medium, which affects the lens's ability to bend light. Understanding this concept is crucial for various applications, including optical instrument design and underwater photography.

Refractive Index and Lensmaker's Equation

The refractive index of a medium is a measure of how much the medium slows down light compared to a vacuum. The lensmaker's equation relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces:

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

where:

• ( f ) is the focal length of the lens
• ( n ) is the refractive index of the lens material relative to the surrounding medium
• ( R_1 ) and ( R_2 ) are the radii of curvature of the lens surfaces

Change in Focal Length

When a lens is placed in a medium other than air, its effective refractive index changes. The effective refractive index ( n' ) is given by:

[ n' = \frac{n_{lens}}{n_{medium}} ]

where:

• ( n_{lens} ) is the refractive index of the lens material
• ( n_{medium} ) is the refractive index of the surrounding medium

The new focal length ( f' ) can be found by substituting ( n' ) into the lensmaker's equation:

[ \frac{1}{f'} = (n' - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) ]

Table of Differences and Important Points

Property	Lens in Air	Lens in Medium
Refractive Index of Surrounding Medium	( n_{air} \approx 1 )	( n_{medium} > 1 ) (usually)
Effective Refractive Index of Lens	( n_{lens} )	( \frac{n_{lens}}{n_{medium}} )
Focal Length	( f )	( f' )
Lensmaker's Equation	( \frac{1}{f} = (n_{lens} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) )	( \frac{1}{f'} = \left( \frac{n_{lens}}{n_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) )

Examples

Example 1: Lens in Water

Consider a lens with a refractive index of 1.5 placed in water with a refractive index of 1.33. The effective refractive index of the lens in water is:

[ n' = \frac{1.5}{1.33} \approx 1.13 ]

If the lens had a focal length of 20 cm in air, its new focal length in water would be:

[ \frac{1}{f'} = (1.13 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) ]

Assuming the lens is thin and ( R_1 = R_2 ), we can simplify the equation to:

[ \frac{1}{f'} = 0.13 \left( \frac{2}{R_1} \right) ]

If ( R_1 = 30 ) cm, then:

[ \frac{1}{f'} = 0.13 \left( \frac{2}{30} \right) = \frac{0.13}{15} ]

[ f' = \frac{15}{0.13} \approx 115.38 \text{ cm} ]

The focal length of the lens in water is significantly longer than in air.

Example 2: Lens in Glass

Suppose a lens with a refractive index of 1.5 is placed in glass with a refractive index of 1.52. The effective refractive index of the lens in glass is:

[ n' = \frac{1.5}{1.52} \approx 0.99 ]

Since ( n' ) is less than 1, the lens will not converge light in the glass and will instead act as a diverging lens. The focal length in this case is negative, indicating a virtual focus.

Conclusion

The focal length of a lens changes when placed in a different medium due to the change in the effective refractive index. This can have significant implications for the design and use of optical systems in various environments. Understanding the relationship between the refractive indices and the focal length is essential for predicting the behavior of lenses under different conditions.