Question

58 through 67 61 59 Lenses with given radii. Object$\mathit{O}$stands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance$\mathit{p}$ , index of refraction n of the lens, radius $\mathit{r}\mathbf{2}$of the nearer lens surface, and radius of the farther lens surface. (All distances are incentimeters.) Find (a) the image distance and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real (R) or virtual $\left(V\right)$, (d) inverted$\mathbf{\left(}\mathit{I}\mathbf{\right)}$ from object or noninverted (NI), and (e) on the same side of the lens as objector on the opposite side

Short answer

a) The image distance $i=-63\mathrm{cm}$

b) The lateral magnification of the object is role="math" localid="1663064780321" $m=+2.2$

c) The image is virtual $\left(V\right)$.

d) The image is non-inverted from object $\left(l\right)$.

e)The image on the same side of the object.

Step-by-step solution

Listing the given quantities

The object distance is$P=+29\mathrm{cm}$

The index of refraction of the lens is$n=1.65$

The radius of the nearer lens surface is$r_1=+35\mathrm{cm}$

The radius of the farther lens surface is $r_2=\infty$.

Understanding the concepts of lens equation and the formula for magnification

We can use the concept of the Lens formula and Lens marker’s equation. The focal length of the lens is positive for a converging lens and negative for a diverging lens. The converging lens can form a virtual as well as real image. If the object is outside the focal point, then it is a real image, and if the object is inside the focal point, then it is a virtual image.

Formula:

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

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

$m=-\frac{i}{P}$

(a) Calculations of the image distance

According to the lens marker’s equation, the expression of the focal length of the lens in air is

$$\begin{gathered} \frac{1}{f}=\left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \\ =\left(1.65-1\right)\left(\frac{1}{35cm}-\frac{1}{\infty }\right) \\ =1.9\times {10}^{-2}cm \end{gathered}$$

$f=+54\mathrm{cm}$

The given lens is a converging lens because the focal length ispositive

For an object in front of the lens, the object distance $P$ and image distance $i$ are related to the lens’s focal length.

$$\begin{gathered} \frac{1}{f}=\frac{1}{P}+\frac{1}{i} \\ \frac{1}{i}=\frac{1}{f}-\frac{1}{P} \end{gathered}$$

$$\begin{gathered} i=\frac{Pf}{P-f} \\ =\frac{\left(+29cm\right)\left(+54cm\right)}{\left(+29cm\right)-\left(+54cm\right)} \\ =-63cm \end{gathered}$$

The image distance $i=-63\mathrm{cm}$

(b) Calculations of the magnification

The lateral magnification is the ratio of the object distance $P$ to the image distance $i$. It is given by

$$\begin{gathered} m=-\frac{i}{P} \\ =-\frac{\left(-63cm\right)}{+29cm} \\ =+2.2 \end{gathered}$$

The lateral magnification of the object is $m=+2.2$

(c) Explanation

Whetherthe image is real $\left(R\right)$or virtual $\left(V\right)$:

If the object is inside the focal point, then it is a virtual image. The image distance is also negative; hence the image is virtual $\left(V\right)$.

(d) Explanation

Whether the image is inverted from object$\left(I\right)$or not inverted$\left(NI\right)$ :

The value of magnification is positive; hence the image is not inverted $\left(NI\right)$.

(e) Explanation

The position of the image:

The value image is negative; hence the image is on the same side as the object.