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MODEL PAPER 6 — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMODEL PAPER 65 Marks
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.
✓

Answer

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

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