Respuesta :
Answer:
Maximum Power = 144.3 D
The associated focal length of the lens = [tex]6.92*10^{-3} m[/tex]
Explanation:
According to the Lens maker's Formula:
[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{R_2} )[/tex]
where;
[tex]n_1[/tex] = the refractive index of the medium
[tex]R_1[/tex] and [tex]R_2[/tex] = radius of curvature on each surface
For a convex lens, The radius of curvature in the front surface will be positive and that of the second surface will be negative . Therefore;
[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{-R_2} ) \\ \\ \frac{1}{f} = (n-1) (\frac{1}{R_1}+\frac{1}{R_2} )[/tex]
At maximum power
[tex]\frac{1}{f} = (1.430-1) (\frac{1}{6.50 \ mm}-\frac{1}{5.50 \ mm} )[/tex]
= [tex]0.144 \ mm^{-1}[/tex]
This Implies
[tex]f = 6.92 mm\\f = 6.92*10^{-3} \ m[/tex]
Therefore; the power is given by the formula:
[tex]P_{max} = \frac{1}{f}[/tex]
[tex]P_{max}= \frac{1}{6.92*10^{-3}}[/tex]
= 144.3 D
The maximum power is 144.3 D and the associated focal length of the lens is 6.92 [tex]\rm \times 10^{-3}[/tex] m and this can be determined by using the lens maker's formula.
Given :
- A typical human lens has an index of refraction of 1.430.
- The lens has a double convex shape, but its curvature can be varied by the ciliary muscles acting around its rim.
- At minimum power, the radius of the front of the lens is 10.0 mm, whereas that of the back is 6.00 mm.
- At maximum power, the radii are 6.50 mm and 5.50 mm.
The lens maker's formula can be used in order to determine the maximum power and associated focal length of the lens.
The lens maker's formula is given below:
[tex]\rm \dfrac{1}{f}=(n - 1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)[/tex]
where f is the focal length of the lens, n is the refractive index, and [tex]\rm R_1[/tex] and [tex]\rm R_2[/tex] are the radius of curvature on each surface.
The radius of curvature of the first surface is positive and the radius of curvature of the second surface is negative. So, the lens maker's formula becomes:
[tex]\rm \dfrac{1}{f}=(n - 1)\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)[/tex]
Now, substitute the values of the known terms in the above formula.
[tex]\rm \dfrac{1}{f}=(1.430 - 1)\left(\dfrac{1}{6.50}+\dfrac{1}{5.50}\right)[/tex]
Now, simplify the above equation in order to determine the value of 'f'.
[tex]\rm \dfrac{1}{f}=0.144\;mm^{-1}[/tex]
f = 6.92 mm = 6.92 [tex]\rm \times 10^{-3}[/tex] m
Now, the maximum power is given by the formula:
[tex]\rm P_{max} =\dfrac{1}{f}[/tex]
[tex]\rm P_{max} = 144.3 \;D[/tex]
For more information, refer to the link given below:
https://brainly.com/question/6839343
