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Ray Optics and Optical Instruments — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsRay Optics and Optical Instruments4 Marks
Question
A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the object are situated at the least distance of distinct vision from the eye. It can be given that: $m = m_e \times m_0,$ where $m_e$ is magnification produced by eye lens and m0 is magnification produced by objective lens.
Consider a compound microscope that consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm.
  1. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will be:
  1. 3.45cm
  2. 5cm
  3. 1.29cm
  4. 2.59cm
  1. How far from the objective should an object be placed in order to obtain the condition described in part (i)?
  1. 4.5cm
  2. 2.5cm
  3. 1.5cm
  4. 3.0cm
  1. What is the magnifying power of the microscope in case of least distinct vision?
  1. 20
  2. 30
  3. 40
  4. 10
  1. The intermediate image formed by the objective of a compound microscope is:
  1. Real, inverted and magnified.
  2. Real, erect, and magnified.
  3. Virtual, erect and magnified.
  4. Virtual, inverted and magnified.
  1. The magnifying power of a compound microscope increases with:
  1. The focal length of objective lens is increased and that of eye lens is decreased.
  2. The focal length of eye lens is increased and that of objective lens is decreased.
  3. Focal lengths of both objects and eye-piece are increased.
  4. Focal lengths of both objects and eye-piece are decreased.
✓

Answer

  1. (b) 5cm
Explanation:
Here, $f_0 = 2.0, f_e = 6.25cm, u_0 = ?$
When the final image is obtained at the least distance of distinct vision:
$v_e = -25cm$
As $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
$\therefore\frac{1}{\text{u}_\text{e}}\frac{1}{\text{v}_\text{e}}\frac{1}{\text{f}_\text{e}}=\frac{1}{-25}-\frac{1}{6.25}$
$=\frac{-1-4}{25}=\frac{-5}{25}=-\frac{1}{5}$
or $\text{u}_\text{e}=-5\text{cm}$
  1. (b) 2.5cm
Explanation:
Distance between objective and eye-piece = 15cm
$\therefore$ Distance of the image from objective is $\text{v}_0=15-5=10\text{cm}$
$\therefore\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{\text{10}}-\frac{1}{\text{2}}=\frac{1-5}{\text{10}}=-\frac{2}{5}$
or $\text{u}_0=-\frac{5}{2}=-2.5\text{cm}$
$\therefore$ Distance of object from objective = 2.5cm
  1. (a) 20
Explanation:
Magnifying power,
$\text{m}=\text{m}_0\times\text{m}_\text{e}=\frac{\text{v}_0}{\text{u}_0}(1+\frac{\text{D}}{\text{fe}})=\frac{10}{2.5}(1+\frac{25}{6.25})=20$
  1. (a) Real, inverted and magnified.
Explanation:
The intermediate image formed by the objective of a compound microscope is real, inverted and magnified.
  1. (d) Focal lengths of both objects and eye-piece are decreased.

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