A homemade vernier uses two identical inclined strips as shown, with inclination angle $θ$ . The projected spacing on the horizontal base is $l$ . For this numerical version, take $l = 1.0 mm$ and $θ = 6 0^{\circ}$ . Find the least count in mm.

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TestHub · Accepted Answer

The horizontal separation of neighbouring vertical marks is $l$ .

Their separation along the inclined edge is $s$ , with $s cos θ = l$ .

$s = \frac{l}{cos θ}$ .

The rotated identical strip presents spacing $l$ along the shared edge.

$LC = s - l = l (\frac{1 - cos θ}{cos θ})$ .

$LC = 1.0 (\frac{1 - 0.5}{0.5}) = 1.0 mm$ .

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