A compound microscope consists of an objective lens of focal length 1 cm and an eyepiece of focal length 5 cm with a separation of 10 cm. The distance between an object and the objective lens, at which the strain on the eye is minimum is n \over 40 cm. The value of n is \underline\hspace2cm .

Geometrical Optics PYQ — JEE Main 2020 · 5 Sep · Shift 1 · Q42 with Solution

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Correct answer: 50

Condition for minimum strain

For a compound microscope, minimum strain on the eye means the final image is formed at infinity.

So the intermediate image formed by the objective must lie at the first focal plane of the eyepiece.

Find image distance for the objective

Given:

Focal length of objective: $f_o = 1\,\text{cm}$
Focal length of eyepiece: $f_e = 5\,\text{cm}$
Separation between lenses: $L = 10\,\text{cm}$

If the final image is at infinity, then the object for the eyepiece must be at its focal distance. Hence the intermediate image must be at a distance $f_e = 5$ cm to the left of the eyepiece.

Since the lenses are separated by $10$ cm, the image formed by the objective is at distance

v_o = L - f_e = 10 - 5 = 5\,\text{cm}

from the objective.

So for the objective lens:

image distance $v = 5$ cm
focal length $f = 1$ cm

Use lens formula for the objective

Using the lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

with sign convention for a convex lens:

$f = +1$ cm
$v = +5$ cm
$u$ will be negative

Thus,

\frac{1}{1} = \frac{1}{5} - \frac{1}{u}

1 = \frac{1}{5} - \frac{1}{u}

-\frac{1}{u} = 1 - \frac{1}{5} = \frac{4}{5}

u = -\frac{5}{4}\,\text{cm}

Therefore the distance between the object and objective lens is

|u| = \frac{5}{4}\,\text{cm}

Match with the given form

Given,

\frac{n}{40} = \frac{5}{4}

So,

n = 40 \times \frac{5}{4} = 50

Final answer

\boxed{n = 50}

The derived answer matches the stored correct answer.

More from Geometrical Optics

View written solutionFree

Correct answer: 50

Condition for minimum strain

For a compound microscope, minimum strain on the eye means the final image is formed at infinity.

So the intermediate image formed by the objective must lie at the first focal plane of the eyepiece.

Find image distance for the objective

Given:

Focal length of objective: $f_o = 1\,\text{cm}$
Focal length of eyepiece: $f_e = 5\,\text{cm}$
Separation between lenses: $L = 10\,\text{cm}$

If the final image is at infinity, then the object for the eyepiece must be at its focal distance. Hence the intermediate image must be at a distance $f_e = 5$ cm to the left of the eyepiece.

Since the lenses are separated by $10$ cm, the image formed by the objective is at distance

v_o = L - f_e = 10 - 5 = 5\,\text{cm}

from the objective.

So for the objective lens:

image distance $v = 5$ cm
focal length $f = 1$ cm

Use lens formula for the objective

Using the lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

with sign convention for a convex lens:

$f = +1$ cm
$v = +5$ cm
$u$ will be negative

Thus,

\frac{1}{1} = \frac{1}{5} - \frac{1}{u}

1 = \frac{1}{5} - \frac{1}{u}

-\frac{1}{u} = 1 - \frac{1}{5} = \frac{4}{5}

u = -\frac{5}{4}\,\text{cm}

Therefore the distance between the object and objective lens is

|u| = \frac{5}{4}\,\text{cm}

Match with the given form

Given,

\frac{n}{40} = \frac{5}{4}

So,

n = 40 \times \frac{5}{4} = 50

Final answer

\boxed{n = 50}

The derived answer matches the stored correct answer.

Geometrical Optics question

Geometrical Optics question

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Geometrical Optics question

Geometrical Optics question

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