Question: Q. 4. (i) Draw a ray diagram showing the image formation by a compound microscope. Obtain expression for total magnification when the image is formed at infinity.

(ii) How does the resolving power of a compound microscope get affected, when

(a) focal length of the objective is decreased.

(b) the wavelength of light is increased?

Give reasons to justify your answer.

U [O.D. I, II, III 2015]

Show Answer

Solution:

Ans. (i) (a)

(ii) Magnification by objectivelens $=\frac{\tan \beta}{\tan \alpha}$

$\frac{h^{\prime}}{h}=\frac{L}{f_{o}}$ (where, the distance between the second

focal point of the objective and the first focal point of the eyepiece is called the tube length of the compound microscope and is denoted by $L$ )

Eyepiece will act as simple microscope, hence we may use the formula of magnification by simple microscope for normal adjustment.

$$ m_{e}=\frac{D}{f_{e}} $$

Total magnification, $\quad m=m_{0} \times m_{e}$

$$ \begin{align*} & =\frac{L}{f_{0}} \times \frac{D^{e}}{f_{e}} \tag{1}\ d_{\text {min }} & =\frac{1.22 f \lambda}{D} \end{align*} $$

(a) From the equation, it is clear that resolving power increases when the focal length of the objective is decreased. This is because the minimum separation, $d_{\text {min }}$ decreases when $f$ is decreased. (b) Resolving power decreases when the wavelength of light is increased. This is because the minimum separation, $d_{\min }$ increases when $\lambda$ is increased. $\mathbf{1}$

[A] Q. 5. (i) Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.

(ii) You are given three lenses of power $0.5 \mathrm{D}, 4 \mathrm{D}$ and $10 \mathrm{D}$ to design a telescope.

(a) Which lenses should be used as objective and eyepiece? Justify your answer.

(b) Why is the aperture of the objective preferred to be large?

A [O.D. I, II, III 2016]

Ans. (i) Try yourself, Similar to Q. 1 (i), Long Answer Type Questions

Definition : It is the ratio of the angle subtended at the eye, by the final image, to the angle which the object subtends at the lens, or the eye.

1

(ii) (a)

Objective $=0.5 \mathrm{D}$

Eye lens $=10 \mathrm{D}$

This choice-would give higher magnification as

$$ \mathrm{M}=\frac{f_{o}}{f_{e}}=\frac{P_{e}}{P_{o}} $$

(b) High resolving power/brighter image/lower limit of resolution

(Any one) 2

[CBSE Marking Scheme 2016]

Commonly Made Error

  • Many students start correctly but get confused in between.
  • Few students derived the expression for relaxed eye.


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