Q55P - Fundamentals Of Physics

Question

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  $\mathbf{\left(}\mathit{R}\mathbf{\right)}$ or virtual $\mathbf{\left(}\mathit{V}\mathbf{\right)}$ , (d) inverted $\mathbf{\left(}\mathit{I}\mathbf{\right)}$  from object $\mathit{O}$ or noninverted  $\left(NI\right)$, and (e) on the same side of the lens as object $\mathit{O}$ or on the opposite side.

 

Step-by-Step Solution

Answer

(a)The image distance   $i=-86\mathrm{cm}$.

(b)The lateral magnification of the object is   $m=+039$

(c)The image is virtual.

(d)The image is not inverted.

(e)The image is on the same side as the object.

Step 1: Listing the given quantities

The object distance is  $P=+2\mathrm{cm}$

The given lens is diverging $\left(D\right)$

The distance between a focal point and the lens is $f=14\mathrm{cm}$.

Step 2: Understanding the concepts of lens equation and the formula for magnification

We can use the Len’s formula. The diverging lens can only form a virtual image.

Formula:

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

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

Step 3: (a) Calculations of the image distance

The given lens is a diverging lens, and thus the focal length value should be negative. 

 $f=-14\mathrm{cm}$

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

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

$$\begin{gathered} i=\frac{Pf}{P-f} \\ =\frac{\left(+22 cm\right)\left(-14 cm\right)}{\left(+22 cm\right)-\left(-14 cm\right)} \\ =-8.6 cm \end{gathered}$$

The image distance  $i=-86\mathrm{cm}$.

Step 4: (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

$$\begin{gathered} m=-\frac{i}{P} \\ =-\frac{\left(-8.6 cm\right)}{+22 cm} \\ =+0.39 \end{gathered}$$

The lateral magnification of the object is   $m=+039$

 

Step 5: (c) Determine whether the image is real or virtual

Whether the image is real $R$  or virtual $V$ :

The image distance is negative; hence the image is virtual.

Step 6: (d) Determine whether the image is inverted or non-inverted

Whether the image is inverted from object or non -inverted 

The value of magnification is positive; hence the image is not inverted $NI$.

Step 7: (e) Determine whether the image is on the same side of the lens as object or on the opposite side

The position of the image:

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