91Ó°ÊÓ

Consider that the focal length of a thin converging lens is 0.3m. A sheet is kept 40m away to display the real image of small tree of height 2m.

Give the formula for magnification in terms of height and distance.

$\mathit{M}\mathbf{=}\frac{{\mathbf{h}}_{\mathbf{2}}}{{\mathbf{h}}_{\mathbf{1}}}$ …â¶Ä¦(1)

$\mathit{M}\mathbf{=}\frac{{\mathbf{d}}_{\mathbf{2}}}{{\mathbf{d}}_{\mathbf{1}}}$ …â¶Ä¦(2)

Give the lens equation.

$\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{d}}_{\mathbf{2}}}$ …â¶Ä¦(3)

Consider the magnification formula in terms of height from Equation (1),

$\mathit{M}\mathbf{=}\frac{{\mathbf{h}}_{\mathbf{2}}}{{\mathbf{h}}_{\mathbf{1}}}$

Consider the magnification formula in terms of distance from Equation (2),

$\mathit{M}\mathbf{=}\frac{{\mathbf{d}}_{\mathbf{2}}}{{\mathbf{d}}_{\mathbf{1}}}$

Compare the Equations (1) and (2),

$\frac{h_2}{h_1}=\frac{d_2}{d_1}$ …â¶Ä¦(4)

Solve the Equation (3) for $d_2$as follows,

$d_2=\frac{f{d}_1}{d_1-f}$ …â¶Ä¦(5)

Substitute the values $f=0.3\text{m}$and $d_1=40\text{m}$in Equation (5),

$d_2=\frac{\left(0.3m\right)\left(40\text{m}\right)}{40\text{m}-\text{0.3m}}$

$d_2=0.302\text{m}$

Solve the Equation (4) for $h_2$ as follows,

$h_2=-\frac{d_2}{d_1}{h}_1$ …â¶Ä¦(6)

Substitute the values $d_2=0.302\text{m}$, $d_1=40\text{m}$and $h_1=2\text{m}$ into the Equation (6).

$$\begin{gathered} h_2=-\frac{0.302\text{m}}{40\text{m}}\left(2\text{m}\right) \\ {h}_2=-0.015\text{m} \end{gathered}$$

The negative sign represents that the height of the image is decreased.

The magnification of the image is,

$M=-\frac{d_2}{d_1}$ …â¶Ä¦(7)

Substitute the values $d_2=0.302\text{m}$,$d_1=40\text{m}$into the Equation (7)

$$\begin{gathered} M=-\frac{0.302\text{m}}{40\text{m}} \\ M=-0.008 \end{gathered}$$

Hence, the image is inverted.

Therefore, The paper must be placed at the distance of 0.302m. The height of image of the tree is 0.0015m. The image of the tree is inverted