When the final image is formed at the least d | vedclass

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(B) For a simple microscope,the lens is a convex lens. The object is placed between the optical center and the focus.Using the lens formula: $\frac{1}

Vedclass · Accepted Answer

$1+\frac{D}{f}$

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(B) For a simple microscope,the lens is a convex lens. The object is placed between the optical center and the focus.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Here,the final image is formed at the least distance of distinct vision,so $v = -D$ (using sign convention).Substituting this into the lens formula: $\frac{1}{-D} - \frac{1}{u} = \frac{1}{f}$.Rearranging for $\frac{1}{u}$: $\frac{1}{u} = -\frac{1}{D} - \frac{1}{f} = -\left(\frac{f+D}{fD}ight)$.The magnification $m$ is given by $m = \frac{v}{u}$.Substituting $v = -D$ and $\frac{1}{u} = -\left(\frac{f+D}{fD}ight)$:$m = (-D) 	imes \left[ -\left(\frac{f+D}{fD}ight) ight] = \frac{D(f+D)}{fD} = \frac{f+D}{f} = 1 + \frac{D}{f}$.

When the final image is formed at the least distance of distinct vision $(D)$ from the eye,the magnifying power of a simple microscope is ($f$ is the focal length of the lens):

Similar Questions

In a microscope, the objective has a focal length $f_0 = 2 \ cm$ and the eyepiece has a focal length $f_e = 4 \ cm$. The tube length is $32 \ cm$. The magnification produced by this microscope for normal adjustment is . . . . . . .

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