Question 89: Ray Optics

Question

A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification $m_1$ when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to $m_2$. The value of $\left|\frac{m_1}{m_2}\right|$ is $\_\_\_\_$ .

Accepted Answer

The focal length of convex lens $=\mathrm{f}_1=+5 \mathrm{~cm}$
The focal length of concave lens $=\mathrm{f}_2=-4 \mathrm{~cm}$
Initial object distance: $u=-10 \mathrm{~cm}$ with respect to convex lens.
Using lens formula for first lens,
$\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1} \Rightarrow \frac{1}{v_1}-\frac{1}{-10}=\frac{1}{5}$
$\Rightarrow \frac{1}{v_1}=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}$
$$ \mathrm{v}_1=10 \mathrm{~cm} $$
So, the image by first lens is formed at a distance of 10 cm from the first lens. The positive sign shows that the first image is formed right of first lens.
Case 1 :
when lenses are combined together without any gap ( $\mathrm{d}=0$ )
When an object is placed at distance $u$, the first lens forms an image at distance $v_1$.
Using lens formula:
$$ \frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}_1} \ldots 	ext { (i) } $$
This image acts as a virtual object for the second lens. Since there is no gap, the object distance for the second lens is simply $\mathrm{v}_1$. It forms the final image at distance v .
$$ \frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2} \ldots (ii)$$
From both the equations,
$\left(\frac{1}{v_1}-\frac{1}{u}ight)+\left(\frac{1}{v}-\frac{1}{v_1}ight)=\frac{1}{f_1}+\frac{1}{f_2}$
$$\Rightarrow $$ $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f_{e q}}$
$$\Rightarrow $$ $\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}$
$$\Rightarrow $$ $\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{5}+\frac{1}{-4}=\frac{4-5}{20}=-\frac{1}{20}$
$$\Rightarrow $$ $$ \mathrm{f}_{\mathrm{eq}}=-20 \mathrm{~cm} $$
Using the lens formula for equivalent lens :
$$ \frac{1}{v}-\frac{1}{u}=\frac{1}{f_{e q}} $$
$$\Rightarrow $$ $\frac{1}{v}-\frac{1}{-10}=-\frac{1}{20}$
$$\Rightarrow $$$\frac{1}{v}=-\frac{1}{20}-\frac{1}{10}$
$$\Rightarrow $$ $\frac{1}{v}=\frac{-1-2}{20}=-\frac{3}{20}$
$$\Rightarrow $$ $$ \mathrm{v}=-\frac{20}{3} \mathrm{~cm} $$
The magnification $\mathrm{m}_1$ for the combined system is the ratio of image distance to object distance:
$$ m_1=\frac{v}{u}=\frac{-20 / 3}{-10}=\frac{2}{3} $$
Case 2 :
When lenses are separated by a gap of d = 1 cm
Refraction through the first lens (Convex, $\mathrm{f}_1=+5 \mathrm{~cm}$ )
The object is still at $\mathrm{u}_1=-10 \mathrm{~cm}$.
$$ \frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1} $$
$$\Rightarrow $$ $\frac{1}{v_1}-\frac{1}{-10}=\frac{1}{5}$
$$\Rightarrow $$ $\frac{1}{v_1}+\frac{1}{10}=\frac{1}{5}$
$$\Rightarrow $$ $\frac{1}{v_1}=\frac{1}{5}-\frac{1}{10}=\frac{2-1}{10}=\frac{1}{10}$
$$\Rightarrow $$ $$ \mathrm{v}_1=+10 \mathrm{~cm} $$
The lateral magnification of the first lens is :
$$ \mathrm{m}_{\mathrm{L} 1}=\frac{\mathrm{v}_1}{\mathrm{u}_1}=\frac{10}{-10}=-1 $$
Refraction through the second lens (Concave, $\mathrm{f}_2=-4 \mathrm{~cm}$ )
The image formed by the first lens acts as the object for the second lens. Since the first image is 10 cm to the right of $L_1$, and $L_2$ is placed 1 cm to the right of $L_1$, the distance of this image from $L_2$ is:
$$ \mathrm{u}_2=\mathrm{v}_1-\mathrm{d}=10-1=+9 \mathrm{~cm} $$
It is a virtual object for the second lens, hence the positive sign.
Now, applying the lens formula for $\mathrm{L}_2$ :
$$ \frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2}=\frac{1}{\mathrm{f}_2} $$
$$\Rightarrow $$ $\frac{1}{v_2}-\frac{1}{9}=\frac{1}{-4}$
$$\Rightarrow $$ $\frac{1}{v_2}=-\frac{1}{4}+\frac{1}{9}=\frac{-9+4}{36}=-\frac{5}{36}$
$$\Rightarrow $$ $$ \mathrm{v}_2=-\frac{36}{5} \mathrm{~cm} $$
The lateral magnification of the second lens is:
$$ \mathrm{m}_{\mathrm{L} 2}=\frac{\mathrm{v}_2}{\mathrm{u}_2}=\frac{-36 / 5}{9}=-\frac{4}{5} $$
So, the total magnification in second case is
$$ \mathrm{m}_2=\mathrm{m}_{\mathrm{L} 1} 	imes \mathrm{m}_{\mathrm{L} 2}=(-1) 	imes\left(-\frac{4}{5}ight)=\frac{4}{5} $$
So, the required ratio of magnification is $\left|\frac{\mathrm{m}_1}{\mathrm{~m}_2}ight|=\frac{2 / 3}{4 / 5}=\frac{5}{6}$.

Answer

$\frac{25}{27}$

Answer

$\frac{5}{6}$

Answer

$\frac{5}{27}$

Answer

$\frac{3}{2}$

Ray Optics

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