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A travelling microscope is used to determine the refractive index of a glass slab. If $40$ divisions are there in $1 \; cm$ on main scale and $50$ Vernier scale divisions are equal to $49$ main scale divisions, then least count of the travelling microscope is $\dots \; \times 10^{-6} \; m$
A
$2$
B
$3$
C
$4$
D
$5$
(JEE MAIN-2022)
Solution
$50 \; VSD =49 \; MSD$
$1 \; VSD =\frac{49}{50} \; MSD$
Least count $=1 MSD -1 VSD$
$=\left(1-\frac{49}{50}\right) MSD =\frac{1}{50} \; MSD$
$1 \; MSD =\frac{1}{40} \; cm$
Least count $=\frac{1}{50 \times 40} \; cm$
$=\frac{1}{2000} cm =\frac{1}{2} \times 10^{-5} \; m$
$=0.5 \times 10^{-5} \; m$
$=5 \times 10^{-6} \; m$
Standard 11
Physics