i. A coin is placed inside a denser medium. Why does it appear to… - Vidyadip

Question

i. A coin is placed inside a denser medium. Why does it appear to be raised? Obtain an expression for the height through which the object appears to be raised in terms of refractive index of the medium and real depth.
ii. A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at the least distance of distinct vision (25 cm)? Calculate the magnifying power of the microscope.

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Answer

i. Due to refraction of light

$\begin{array}{l}\text { In } \triangle OAB \\
\sin i =\frac{A B}{O B}\end{array}$
In $\triangle IAB , \sin r =\frac{A B}{I B}$
According to Snell's law
$\begin{array}{l}\frac{1}{\mu}=\frac{\sin i}{\sin r}=\frac{I B}{O B} \text { When angles are small, } OB \approx OA \text { and } IB \approx IA \\
\mu=\frac{O A}{I A}=\frac{x}{y}\end{array}$
Height through which object is raised = x - y
$\begin{array}{l}= x -\frac{x}{\mu} \\
= x \left(1-\frac{1}{\mu}\right)\end{array}$
ii. $f _0=2 cm$
$\begin{array}{l} f _{ e }=6.25 cm \\
L = v _0+\left| u _{ e }\right|=15 cm \\
v _{ e }=-25 cm\end{array}$
$\begin{array}{l}\frac{1}{V_e}-\frac{1}{u_e}=\frac{1}{f_e} \\
\frac{1}{-25}-\frac{1}{u_e}=\frac{1}{6.25} \\
u _{ e }=-5 cm\end{array}$
Now, $L=v_0+|-5|=15 cm$
$V_o=10 cm$
$\begin{array}{l}\text { Now, } \frac{1}{f_o}=\frac{1}{V_o}-\frac{1}{u_o} \\
u _0=2.5 cm \\
MP =\frac{V_o}{u_o}\left[1+\frac{D}{f_o}\right] \\
=-20\end{array}$

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