3 Questions around this concept.
The eye-piece and objective of a microscope, having focal lengths of 0.3 m and 0.4 m respectively, are separated by a distance of 0.2 m. Now the eyepiece and the objective are to be interchanged such that the angular magnification of the instrument remains the same. The new separation between the lenses is:
The focal length of the object glass of a microscope is 2 cm, that of the eye-piece is 4 cm and the distance between them is 20 cm. What should be the object distance from the object-glass if the final image is to be seen at least a distance of distinct vision (25 cm) from the eyepiece? What will be the magnification in this situation?
Magnifying power: It is defined as the ratio between the dimensions of the image and the object.
Magnifying power of an optical instrument is given by,
$m=\frac{\text { Visual angle with instrument }(\beta)}{\text { Visual angle when object is placed at least distance of distinct vision }(\alpha)}$

Simple Microscope
Case 1: Magnification's, when the final image is formed at D and $\infty$.
$$
\begin{aligned}
& \text { (i.e. } m_D \text { and } m_{\infty} \text { ) } \\
& m_D=\left(1+\frac{D}{f}\right)_{\max \text { and }} \\
& m_{\infty}=\left(\frac{D}{f}\right)_{\min }
\end{aligned}
$$
Case 2: If lens is kept at a distance a from the eye then
$$
\begin{aligned}
& m_D=1+\frac{D-a}{f} \text { and } \\
& m_{\infty}=\frac{D-a}{f}
\end{aligned}
$$
"Stay in the loop. Receive exam news, study resources, and expert advice!"
