Question

em 4
what is the power of a 37.5 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 - 6.25 - 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) or \( f = \frac{1}{P} \) (where \( P \) is in diopters and \( f \) is in meters).

Part A:

Step2: Convert focal length to meters

The focal length \( f_1 = 37.5\space cm \). Since \( 1\space m=100\space cm \), we convert \( 37.5\space cm \) to meters: \( f_1=\frac{37.5}{100}=0.375\space m \).

Step3: Calculate the power

Using the formula \( P_1=\frac{1}{f_1} \), substitute \( f_1 = 0.375\space m \): \( P_1=\frac{1}{0.375}\space D \approx 2.666...\space D \). Rounding to three significant figures, \( P_1 = 2.67\space D \).

Part B:

Step2: Use the formula for focal length

Given \( P_2=- 6.25\space D \), use \( f_2=\frac{1}{P_2} \). Substitute \( P_2=-6.25\space D \): \( f_2=\frac{1}{- 6.25}\space m=- 0.16\space m \). But we need to convert to centimeters? Wait, no, the formula gives \( f \) in meters. Wait, \( f_2=\frac{1}{-6.25}\space m=-0.16\space m \)? Wait, no: \( \frac{1}{6.25}=0.16 \), so \( \frac{1}{- 6.25}=- 0.16\space m \)? Wait, no, \( 1\div6.25 = 0.16 \), so \( 1\div(- 6.25)=- 0.16\space m \)? Wait, no, \( 1/6.25 = 0.16 \), so \( f_2=\frac{1}{-6.25}\space m=-0.160\space m \) (three significant figures). Wait, \( 1\div6.25 = 0.16 \), but to three significant figures, \( 1\div6.25 = 0.160\space m \)? Wait, no, \( 6.25 \) has three significant figures, so \( 1/6.25 = 0.160\space m \) (since \( 1.00\div6.25 = 0.160 \) when considering significant figures? Wait, no, the formula is \( f=\frac{1}{P} \), where \( P \) is in diopters, so \( f \) is in meters. So for \( P=-6.25\space D \), \( f_2=\frac{1}{-6.25}\space m=-0.160\space m \) (or \( - 16.0\space cm \), but the question says "include the appropriate units". Wait, the formula gives \( f \) in meters, but we can also express in centimeters. Wait, no, let's recalculate: \( f_2=\frac{1}{P_2}=\frac{1}{-6.25}\space m=-0.160\space m \) (since \( 1\div6.25 = 0.16 \), but with three significant figures, it's \( 0.160\space m \) or \( - 16.0\space cm \)? Wait, no, \( 6.25 \) has three significant figures, so \( 1/6.25 = 0.160\space m \) (because \( 1.00\div6.25 = 0.160 \) when we consider that 1 can be thought of as 1.00 for three significant figures? Wait, no, the number 1 is exact here (from the formula \( f = 1/P \)), so the significant figures come from \( P \). So \( P=-6.25\space D \) (three significant figures), so \( f_2=\frac{1}{-6.25}\space m=-0.160\space m \) (three significant figures) or \( - 16.0\space cm \). Wait, but let's do the calculation: \( 1\div6.25 = 0.16 \), but to three significant figures, \( 0.160\space m \) (since \( 6.25 \) has three, so the result should have three). So \( f_2=-0.160\space m \) or \( - 16.0\space cm \).

Answer:

s:

Part A:

\( P_1=\boxed{2.67}\space D \)

Part B:

\( f_2=\boxed{-0.160}\space m \) (or \( \boxed{-16.0}\space cm \))