You are given three lenses L1, L2 and L3 each of focal length 20 cm. An object is kept at 40 cm in front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations between L1, L2 and L3.


Given, 

f1 = f2 = f3 = 20 cm 

In case of lens L1,

Object distance, u1 = -40 cm

Using Lens formula,

space space space space space 1 over straight v subscript 1 space minus space 1 over straight u subscript 1 space equals space 1 over straight f subscript 1

rightwards double arrow space 1 over straight v subscript 1 space equals space 1 over 20 space minus space minus 1 over 40

rightwards double arrow space space straight v subscript 1 space equals space 40 space cm

For lens L3,

straight f subscript 3 space equals space 20 space cm. space straight v subscript 3 space equals space 20 space cm

Object space distance comma space straight u subscript 3 space equals space ?

Using space lens space formula comma space we space have

1 over straight v subscript 3 minus 1 over straight u subscript 3 space equals space 1 over straight f subscript 3 space

rightwards double arrow space 1 over 20 minus 1 over straight u subscript 3 space equals space 1 over 20

rightwards double arrow space space 1 over straight u subscript 3 space equals 0 space
rightwards double arrow space space space space straight u subscript 3 space equals space infinity 

So, lens L2 should produce the image at infinity. Thus, objective should be at focus. The image formed by lens L1 is at 40 cm on the right side of lens L1 which lies at 20 cm left of lens L2 i.e., focus of lens L2.

Hence, the distance between L1 and L2 = 40 + 20 = 60 cm.

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An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?


Given,

Accelerating voltage for electrons = 50 kV = 50 x 103 V

De-Broglie wavelength is given by, 

 

So, from the above formula, we can say that the resolving power of an electron microscope is much greater than that of an optical microscope.

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For the same value of angle incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum? 


As per Snell’s law we have,

 
For given i, v sin r 
r is minimum in medium A, so the velocity of light is minimum in medium A.



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An object AB is kept in front of a concave mirror as shown in the figure.

(i) Complete the ray diagram showing the image formation of the object.

(ii) How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black?        

(i) Image formed will be inverted diminished between C and F.

 

ii) When the lower half of the mirror’s reflecting surface is painted black, the position of the image and its intensity will get reduced.

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Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope


 

 

 

Advantages of reflecting telescope over refracting telescope:

(i) It is free from chromatic aberration.

(ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror.

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