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State the formula for the power of a lens.

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5 parts
17

B. +5.0 D\text{B. } +5.0 \, \text{D}

18

C. −1 m\text{C. } -1 \, \text{m}

19

A. V=IR\text{A. } V = IR

20

D. increase in length\text{D. increase in length}

21

B. ampere\text{B. ampere}

Step-by-step solution

6 steps

Here are the solutions to the questions:

Part 17· 3 steps

  1. State the formula for the power of a lens.

    The power of a lens PP is the reciprocal of its focal length ff in meters. For a converging lens, the focal length is positive.

    P=1fP = \frac{1}{f}

  2. Convert the focal length to meters.

    Given f=20 cmf = 20 \, \text{cm}.

    f=20 cm=0.20 mf = 20 \, \text{cm} = 0.20 \, \text{m}

  3. Calculate the power of the lens.

    P=10.20 m=5 DP = \frac{1}{0.20 \, \text{m}} = 5 \, \text{D}

    Since it is a converging lens, the power is positive.

    P=+5.0 DP = +5.0 \, \text{D}

    The final answer is B. +5.0 D\boxed{\text{B. } +5.0 \, \text{D}}.

Part 18

C. −1 m\text{C. } -1 \, \text{m}

  1. State the relationship between the focal length and radius of curvature for a spherical mirror.

    The focal length ff of a spherical mirror is half its radius of curvature RR. For a convex mirror, the focal length is negative.

    f=R2f = \frac{R}{2}

  2. Substitute the given radius of curvature.

    Given R=2 mR = 2 \, \text{m}.

    f=2 m2=1 mf = \frac{2 \, \text{m}}{2} = 1 \, \text{m}

  3. Apply the sign convention for a convex mirror.

    For a convex mirror, the focal length is negative.

    f=−1 mf = -1 \, \text{m}

    The final answer is C. −1 m\boxed{\text{C. } -1 \, \text{m}}.

Part 19

A. V=IR\text{A. } V = IR

Ohm's law states that the voltage (VV) across a conductor is directly proportional to the current (II) flowing through it, with the constant of proportionality being the resistance (RR).

V=IRV = IR

The final answer is A. V=IR\boxed{\text{A. } V = IR}.

Part 20

D. increase in length\text{D. increase in length}

The resistance RR of a wire is given by the formula:

R=ρLAR = \rho \frac{L}{A}

where ρ\rho is the resistivity, LL is the length, and AA is the cross-sectional area.
From this formula, resistance is directly proportional to length. Therefore, an increase in length will increase the resistance of the wire.

The final answer is D. increase in length\boxed{\text{D. increase in length}}.

Part 21

B. ampere\text{B. ampere}

The unit of electric current is the ampere.

The final answer is B. ampere\boxed{\text{B. ampere}}.

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