A major breakthrough in the studies of cells came with the development of an electron microscope. This is because:

1. the resolution power of the electron microscope is much higher than that of the light microscope.
2. the resolving power of the electron microscope is 200-350 nm compared to 0.1-0.2 nm for the light microscope.
3. electron beam can pass through thick materials, whereas light microscopy requires thin sections.
4. the electron microscope is more powerful than the light microscope as it uses a beam of electrons that has a wavelength much longer than that of photons.

Subtopic:  Resolving Power of Optical Devices (OLD NCERT) |
 54%
From NCERT
AIPMT - 2006
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In a double-slit experiment, when the light of wavelength \(400~\text{nm}\) was used, the angular width of the first minima formed on a screen placed \(1~\text{m}\) away, was found to be \(0.2^{\circ}.\) What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? \(\left(\mu_{\text{water}} = \dfrac{4}{3}\right)\)
1. \(0.1^{\circ}\)
2. \(0.266^{\circ}\)
3. \(0.15^{\circ}\)
4. \(0.05^{\circ}\)

Subtopic:  Young's Double Slit Experiment |
 76%
From NCERT
NEET - 2019
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Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:

1.  2l1+l2

2.  I1+l22

3.  I1-l22

4.  2I1-l2

Subtopic:  Superposition Principle |
 74%
From NCERT
AIPMT - 2008
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A parallel beam of fast-moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?
1. The angular width of the central maximum of the diffraction pattern will increase.
2. The angular width of the central maximum will decrease.
3. The angular width of the central maximum will be unaffected.
4. A diffraction pattern is not observed on the screen in the case of electrons.
Subtopic:  Diffraction |
 52%
From NCERT
AIPMT - 2013
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For a parallel beam of monochromatic light of wavelength \(\lambda\), diffraction is produced by a single slit whose width \(a\) is much greater than the wavelength of the light. If \(D\) is the distance of the screen from the slit, the width of the central maxima will be:

1. \(\dfrac{2D\lambda}{a}\) 2. \(\dfrac{D\lambda}{a}\)
3. \(\dfrac{Da}{\lambda}\) 4. \(\dfrac{2Da}{\lambda}\)
Subtopic:  Diffraction |
 82%
From NCERT
NEET - 2015
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In Young's double-slit experiment, the separation \(d\) between the slits is \(2~\text{mm}\), the wavelength \(\lambda\) of the light used is \(5896~\mathring{A}\) and distance \(D\) between the screen and slits is \(100~\text{cm}\). It is found that the angular width of the fringes is \(0.20^{\circ}\). To increase the fringe angular width to \(0.21^{\circ}\) (with same \(\lambda\) and \(D\)) the separation between the slits needs to be changed to:
1. \(1.8~\text{mm}\)
2. \(1.9~\text{mm}\)
3. \(2.1~\text{mm}\)
4. \(1.7~\text{mm}\)

Subtopic:  Young's Double Slit Experiment |
 76%
From NCERT
NEET - 2018
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Unpolarised light is incident from the air on a plane surface of a material of refractive index \(\mu\). At a particular angle of incidence \(i\), it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
1. the reflected light is polarised with its electric vector parallel to the plane of incidence.
2. the reflected light is polarised with its electric vector perpendicular to the plane of incidence.
3. \(i = \text{sin}^{-1}\frac{1}{\mu}\)
4. \(i = \text{tan}^{-1}\frac{1}{\mu}\)
Subtopic:  Polarization of Light |
 60%
From NCERT
NEET - 2018
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Young’s double-slit experiment is first performed in air and then in a medium other than air. It is found that the \(8^\mathrm{th}\) bright fringe in the medium lies where the \(5^\mathrm{th}\) dark fringe lies in the air. The refractive index of the medium is nearly:
1. \(1.59\)
2. \(1.69\)
3. \(1.78\)
4. \(1.25\)  
Subtopic:  Young's Double Slit Experiment |
 63%
From NCERT
NEET - 2017
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In Young's double-slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has a path difference:

1. \( \dfrac{5\lambda}{2} \) 2. \( \dfrac{10\lambda}{2} \)
3. \( \dfrac{9\lambda}{2} \) 4. \( \dfrac{11\lambda}{2} \)
Subtopic:  Young's Double Slit Experiment |
 76%
From NCERT
NEET - 2019
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The angular width of the central maximum in the Fraunhofer diffraction for \(\lambda=6000~{\mathring{A}}\) is \(\theta_0\). When the same slit is illuminated by another monochromatic light, the angular width decreases by \(30\%\). The wavelength of this light is:
1. \(1800~{\mathring{A}}\)
2. \(4200~{\mathring{A}}\)
3. \(420~{\mathring{A}}\)
4. \(6000~{\mathring{A}}\)

Subtopic:  Diffraction |
 76%
From NCERT
NEET - 2019
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