Chapter 34: Q140P (page 1046)

Suppose the farthest distance a person can see without visual aid is $50 c m$ . (a) What is the focal length of the corrective lens that will allow the person to see very far away? (b) Is the lens converging or diverging? (c) The power Pof a lens (in diopters) is equal to $1 / f$ , where $f$ is in meters. What is $p$ for the lens?

Short Answer

Expert verified

Focal length of the corrective lens $f = - 0.50 c m$
The lens is diverging
Power of the lens $P = - 2.0 d i o p t e r s$

Step by step solution

Listing the given quantities

Far point of the person $i = 50 c m$

Understanding the concepts of lens formula and focal length

We will use the lens formula to find the focal length of the lens. The reciprocal of the focal length gives the power of the lens. If the focal length of the lens is negative, it is a diverging lens

Formula:

$\frac{1}{f} = \frac{1}{p} + \frac{1}{i}$

Power of the lens,

role="math" localid="1663046890549" $\frac{1}{f} = p$

Calculations of the focal length of the corrective lens

(a)

Let’s take the object distance to be at infinity, i.e., $p = \infty$ .

The image is inverted, therefore,then $i = - 0.50 m$

$\begin{matrix} \frac{1}{f} = \frac{1}{p} + \frac{1}{i} \\ = \frac{1}{\infty} + \frac{1}{- 0.50} \end{matrix}$

Hence, the focal length of the lens is $f = - 0.50 m$

Type of the lens

(b)

As $f < 0$ , therefore the lens is diverging.

Calculations of the power of the lens

(c)

The power of the lens is defined as

$\begin{matrix} P = \frac{1}{f} \\ = \frac{1}{- 0.50} \\ = - 2.0 d i o p t e r s \end{matrix}$

Power of the lensrole="math" localid="1663047184088" $P = - 2.0 d i o p t e r s$

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Short Answer

Step by step solution

Listing the given quantities

Understanding the concepts of lens formula and focal length

Calculations of the focal length of the corrective lens

Type of the lens

Calculations of the power of the lens

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