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QJEEAdv19Ph23d4: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ph23d4 Solution: An optical bench has 1.5 m long scale having four equal…

An optical bench has 1.5 m long scale having four equal divisions in each cm. While measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is ..............

Step-by-Step Verified Solution

u=(x2−x1)=75−45=30u = ({x_2} - {x_1}) = 75 - 45 = 30 cm

Δu=Δx2+Δx1=14+14=12\Delta u = \Delta {x_2} + \Delta {x_1} = {1 \over 4} + {1 \over 4} = {1 \over 2} cm

v=(x3−x2)=135−75=60v = ({x_3} - {x_2}) = 135 - 75 = 60 cm

Δv=Δx3+Δx2=14+14=12\Delta v = \Delta {x_3} + \Delta {x_2} = {1 \over 4} + {1 \over 4} = {1 \over 2} cm

JEE Advanced 2019 Paper 2 Offline Physics - Units & Measurements Question 9 English Explanation

∴\therefore 1v−1u=1f⇒160+130=1f{1 \over v} - {1 \over u} = {1 \over f} \Rightarrow {1 \over {60}} + {1 \over {30}} = {1 \over f}

∴\therefore f = 20 cm Also, −dvv2+−duu2=−dff2{{ - dv} \over {{v^2}}} + {{ - du} \over {{u^2}}} = {{ - df} \over {{f^2}}}

⇒dff=f[dvv2+duu2]=20[1602+1302]12\Rightarrow {{df} \over f} = f\left[ {{{dv} \over {{v^2}}} + {{du} \over {{u^2}}}} \right] = 20\left[ {{1 \over {{{60}^2}}} + {1 \over {{{30}^2}}}} \right]{1 \over 2}

∴\therefore dff×100=10[136+19]=5036{{df} \over f} \times 100 = 10\left[ {{1 \over {36}} + {1 \over 9}} \right] = {{50} \over {36}} = 1.38 and 1.39 (both)

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