A) -4
B) 5
C) 10
D) -10
Correct Answer: D
Solution :
For a compound microscope the magnifying power is given by \[\alpha ={{\omega }^{2}}\theta \] When the final image is at least distance of distance vision then \[\Rightarrow \] So, \[{{\omega }^{2}}=\frac{2qE}{mL}\] \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{mL}{2qE}}\] \[\frac{T}{4}.\] \[\text{t=}\frac{\text{T}}{\text{4}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\sqrt{\frac{\text{mL}}{\text{2qE}}}\] \[\text{250Jk}{{\text{g}}^{\text{-1}}}{{\text{K}}^{\text{-1}}}\] \[\Delta \theta =0.4K\] \[\text{U=}\frac{\text{1}}{\text{2}}\text{C}{{\text{V}}^{\text{2}}}\] The negative sign indicates that the image formed by objective is inverted.
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