Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Geometrical Optics question

2020 · 4 Sep · Shift 1 · Q61
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Geometrical Optics
  5. /2020 · 4 Sep · Shift 1 · Q61

Geometrical Optics question

2020 · 4 Sep · Shift 1 · Q61

JEE MainPhysicsGeometrical OpticsNumerical+4 / −1
In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eye-piece lens (in cm) is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 6.25

  1. Given data
  • Total magnification of compound microscope: M=100M = 100M=100
  • Focal length of objective: fo=1 cmf_o = 1\,\text{cm}fo​=1cm
  • Tube length: L=20 cmL = 20\,\text{cm}L=20cm
  • Final virtual image is formed at the least distance of distinct vision: D=25 cmD = 25\,\text{cm}D=25cm

We need to find focal length of eyepiece, fef_efe​.

  1. Magnifying power of a compound microscope

For a compound microscope in normal adjustment with final image at distance DDD from the eyepiece, the magnifying power is

M=mo meM = m_o\, m_eM=mo​me​

where

  • objective magnification: mo≈Lfom_o \approx \frac{L}{f_o}mo​≈fo​L​
  • eyepiece magnification: me=1+Dfem_e = 1 + \frac{D}{f_e}me​=1+fe​D​

So,

M=Lfo(1+Dfe)M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)M=fo​L​(1+fe​D​)
  1. Substitute the given values
100=201(1+25fe)100 = \frac{20}{1}\left(1 + \frac{25}{f_e}\right)100=120​(1+fe​25​) 100=20(1+25fe)100 = 20\left(1 + \frac{25}{f_e}\right)100=20(1+fe​25​)

Divide by 202020:

5=1+25fe5 = 1 + \frac{25}{f_e}5=1+fe​25​ 25fe=4\frac{25}{f_e} = 4fe​25​=4 fe=254=6.25 cmf_e = \frac{25}{4} = 6.25\,\text{cm}fe​=425​=6.25cm
  1. Final answer

The focal length of the eyepiece is

6.25 cm\boxed{6.25\,\text{cm}}6.25cm​
PreviousNext

More from Geometrical Optics

  • The distance between an object and a screen is 100 cm. A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is 40 cm. If the power…2020 · Numerical
  • 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 40n​…2020 · Numerical
  • For a concave lens of focal length f, the relation between object and image distances u and v, respectively, from its pole can best be represented by (u = v is the reference line) :2020 · MCQ
  • A prism of angle A = 1o has a refractive index μ= 1.5. A good estimate for the minimum angle of deviation (in degrees) is close to 10N​. Value of N is ​.2020 · Numerical
  • A point like object is placed at a distance of 1 m in front of a convex lens of focal length 0.5 m. A plane mirror is placed at a distance of 2 m behind the lens. The position and nature of the final image formed by the system is :2020 · MCQ
  • A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power 1.5 P is :2020 · MCQ
  • If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :2020 · MCQ
  • A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid of refractive index 1.42. If its focal length in liquid is f1 , then the ratio ff1​​ is closest to the integer :2020 · MCQ