Question

50 through 57 55, 57 53 Thin lenses. Object $\mathit{O}$stands on the central axis of a thin symmetric lens. For this situation, each problem in Table 34-6 gives object distance p (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance $\mathit{i}$and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real $\left(R\right)$ or virtual $\mathbf{\left(}\mathit{V}\mathbf{\right)}$ , (d) inverted $\mathbf{\left(}\mathit{I}\mathbf{\right)}$from object $\mathit{O}$ or non inverted $\mathbf{\left(}\mathit{N}\mathit{I}\mathbf{\right)}$ , and (e) on the same side of the lens as object $\mathit{O}$or on the opposite side.

Short answer

a) The image distance $i=+36\mathrm{cm}$

b) The lateral magnification of the object is $i=+36\mathrm{cm}$

c) The image is virtual $V$

d) The image is inverted from object $I$.

e) The image on the opposite side as the object.

Step-by-step solution

Listing the given quantities

The object distance is $P=+45\mathrm{cm}$

The given lens is a converging lens.

The distance between a focal point and the lens is $P=+45\mathrm{cm}$

Understanding the concepts of lens equation and the formula for magnification

We can use the Lens formula. A converging lens can form a virtual as well as a real image. If the object is outside the focal point, it is a real image, and if the object is inside the focal point, it is a virtual image.

Formula:

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

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

(a) Calculations of the image distance

The given lens is a converging lens, and thus the focal length value should be positive.

$f=+20\mathrm{cm}$

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

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

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

$$\begin{gathered} i=\frac{Pf}{P-f} \\ =\frac{\left(+45cm\right)\left(+20cm\right)}{\left(+45cm\right)-\left(+20cm\right)} \\ =+36cm \end{gathered}$$

The image distance $i=+36\mathrm{cm}$.

(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

$$\begin{gathered} m=-\frac{i}{P} \\ =-\frac{\left(+36cm\right)}{+45cm} \\ =-0.80 \end{gathered}$$

The lateral magnification of the object is $m=-080.0$

(c) Explanation

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

It the object is outside the focal point, then it is real image. The image distance is positive; hence the image is real.

(d) Explanation

Whether the image is inverted from object$\left(I\right)$or non -invertedrole="math" localid="1663056876086" $\left(NI\right)$:

The value of magnification is negative; hence the image is inverted $\left(I\right)$.

(e) Explanation

The position of the image:

The value of image is positive; hence the image is on the opposite side as the object.