Search any question & find its solution
Question:
Answered & Verified by Expert
A beam of parallel rays is brought to a focus by a plano-convex lens. A thin concave lens of the same focal length is joined to the first lens. The effect of this is
Options:
Solution:
2884 Upvotes
Verified Answer
The correct answer is:
the focus shifts to infinity
The combined focal length of plano-convex lens is
$$
\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}
$$
where, $f_{1}=\infty$ for the plane surface and $f_{2}=f$
$$
\begin{array}{ll}
\therefore \quad & \frac{1}{f}=\frac{1}{\infty}+\frac{1}{f} \\
\Rightarrow \quad f & =f
\end{array}
$$
Now, when concave lens of same focal length is joined to first lens, then combined focal length
$$
\begin{array}{ll}
\Rightarrow & \frac{1}{F}=\frac{1}{F_{1}}+\frac{1}{F_{2}} \\
\Rightarrow \quad \frac{1}{F} & =\frac{1}{f}-\frac{1}{f} \\
\Rightarrow \quad & \frac{1}{F}=0 \Rightarrow F=\infty
\end{array}
$$
Thus, the image can be focused on infinity or focus shifts to infinity.
$$
\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}
$$
where, $f_{1}=\infty$ for the plane surface and $f_{2}=f$
$$
\begin{array}{ll}
\therefore \quad & \frac{1}{f}=\frac{1}{\infty}+\frac{1}{f} \\
\Rightarrow \quad f & =f
\end{array}
$$
Now, when concave lens of same focal length is joined to first lens, then combined focal length
$$
\begin{array}{ll}
\Rightarrow & \frac{1}{F}=\frac{1}{F_{1}}+\frac{1}{F_{2}} \\
\Rightarrow \quad \frac{1}{F} & =\frac{1}{f}-\frac{1}{f} \\
\Rightarrow \quad & \frac{1}{F}=0 \Rightarrow F=\infty
\end{array}
$$
Thus, the image can be focused on infinity or focus shifts to infinity.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.