Question

what is the power of a 31.5 cm focal-length lens?
express your answer using three significant figures.
p₁ = d
submit request answer
part b
what is the focal length of a -7.75-diopter lens?
express your answer to three significant figures and include the appropriate units.
f₂ = value units

Explanation:

Step1: Recall the formula for lens power

The power \( P \) of a lens (in diopters, D) is given by the reciprocal of the focal length \( f \) (in meters), so \( P=\frac{1}{f} \) (where \( f \) is in meters).

Step2: Solve Part A

First, convert the focal length from centimeters to meters. Given \( f_1 = 31.5\space cm \), since \( 1\space m=100\space cm \), we have \( f_1=\frac{31.5}{100}=0.315\space m \).
Now, calculate the power using \( P_1=\frac{1}{f_1} \). Substitute \( f_1 = 0.315\space m \) into the formula: \( P_1=\frac{1}{0.315}\approx3.17\space D \) (rounded to three significant figures).

Step3: Solve Part B

We know the power \( P_2=-7.75\space D \), and we use the formula \( f_2=\frac{1}{P_2} \).
Substitute \( P_2=-7.75\space D \) into the formula: \( f_2=\frac{1}{-7.75}\approx - 0.129\space m \) (rounded to three significant figures). To express in centimeters (a common unit for focal length), since \( 1\space m = 100\space cm \), \( f_2=-0.129\times100=-12.9\space cm \) (or we can keep it in meters as \( -0.129\space m \), but centimeters is also appropriate here).

Answer:

Part A:

\( P_1 = \boldsymbol{3.17}\space D \)

Part B:

\( f_2 = \boldsymbol{-12.9}\space cm \) (or \( \boldsymbol{-0.129}\space m \))