Answer
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Hint: Calculate the image formed due to the converging lens. This distance when added to the distance between lens and mirror will become the object distance for the converging mirror. Thus, we can find the focal lengths of mirrors easily.
Formula used:
$\begin{align}
& \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\
& \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} \\
\end{align}$
Complete answer:
Converging lens is present first, thus, the rays pass through the lens.
For converging lens, object distance, focal length is given as $u=-2cm,f=5cm$.
Applying the lens formula, we get,
$\begin{align}
& \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\
& \dfrac{1}{5}=\dfrac{1}{v}-\dfrac{1}{-2} \\
& v=-3.33cm. \\
& \\
\end{align}$
Therefore, the image of the object source formed by the converging lens is $3.33cm$ from the lens on the same side of the object.
As the image is formed on the same side of the object, the distance of object for the converging mirror will become, ${{u}_{1}}=40+3.33=43.33cm$
Given, the final beam of ray coming out of the total system is parallel to the principal axis. We know, if the ray is parallel to the principal axis, the image formed is at infinity. The dame is the case here.
Therefore, the image distance and object distance are $\infty ,43.33cm$ respectively.
Applying lens formula to the above values, we get,
$\begin{align}
& \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\
& \dfrac{1}{f}=\dfrac{1}{\infty }-\dfrac{1}{43.33} \\
& f=43.33cm \\
\end{align}$
Therefore, the focal length of the converging mirror is $43.33cm$.
Additional Information:
The lens formula gives us the relation between object distance, image distance and focal length of the lens. The convex lens is also called a converging lens as the ray’s incident on the lens converges after falling on the lens while the concave lens is called diverging lens as the rays diverge. The lens makers formula was discovered by Descartes.
Note:
In the above problem, the image formed due to the converging lens is on the same side of the object, therefore, the object distance for the converging mirror is the sum of distance between lens and image distance. If in case, the image is formed opposite to the object, the object distance for the error is the subtraction of both the values.
Formula used:
$\begin{align}
& \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\
& \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} \\
\end{align}$
Complete answer:
Converging lens is present first, thus, the rays pass through the lens.
For converging lens, object distance, focal length is given as $u=-2cm,f=5cm$.
Applying the lens formula, we get,
$\begin{align}
& \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\
& \dfrac{1}{5}=\dfrac{1}{v}-\dfrac{1}{-2} \\
& v=-3.33cm. \\
& \\
\end{align}$
Therefore, the image of the object source formed by the converging lens is $3.33cm$ from the lens on the same side of the object.
As the image is formed on the same side of the object, the distance of object for the converging mirror will become, ${{u}_{1}}=40+3.33=43.33cm$
Given, the final beam of ray coming out of the total system is parallel to the principal axis. We know, if the ray is parallel to the principal axis, the image formed is at infinity. The dame is the case here.
Therefore, the image distance and object distance are $\infty ,43.33cm$ respectively.
Applying lens formula to the above values, we get,
$\begin{align}
& \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\
& \dfrac{1}{f}=\dfrac{1}{\infty }-\dfrac{1}{43.33} \\
& f=43.33cm \\
\end{align}$
Therefore, the focal length of the converging mirror is $43.33cm$.
Additional Information:
The lens formula gives us the relation between object distance, image distance and focal length of the lens. The convex lens is also called a converging lens as the ray’s incident on the lens converges after falling on the lens while the concave lens is called diverging lens as the rays diverge. The lens makers formula was discovered by Descartes.
Note:
In the above problem, the image formed due to the converging lens is on the same side of the object, therefore, the object distance for the converging mirror is the sum of distance between lens and image distance. If in case, the image is formed opposite to the object, the object distance for the error is the subtraction of both the values.
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