NEET Wave Optics MCQs & PYQs

Practice NEET Wave Optics Questions

Question 151:

easy

A beam of light of $lambda = 600 nm$ from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is:

(2014)

The distance between the first dark fringes on either side of the central bright fringe is the linear width of the central maximum, given by $W = \frac{2D\lambda}{a}$. Substituting the values, $W = \frac{2 \times 2 \times 600 \times 10^{-9}}{1 \times 10^{-3}} = 2400 \times 10^{-6} m = 2.4 mm$.

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Question 152:

easy

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?

(2013)

The de Broglie wavelength of an electron is $\lambda = \frac{h}{mv}$. As the speed $v$ increases, the wavelength $\lambda$ decreases. The angular width of the central maximum is $2\theta = \frac{2\lambda}{a}$. Since $\lambda$ decreases, the angular width will decrease.

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Question 153:

easy

A parallel beam of monochromatic light of wavelength $5000 \AA$ is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on a screen placed in focal plane. The first minimum will be formed for the angle of diffraction equal to

(1993)

For the first minimum in a single slit diffraction pattern, $a \sin\theta = \lambda$. Thus, $\sin\theta = \frac{\lambda}{a} = \frac{5000 \times 10^{-10}}{0.001 \times 10^{-3}} = \frac{5 \times 10^{-7}}{10^{-6}} = 0.5$. Therefore, $\theta = \sin^{-1}(0.5) = 30^\circ$.

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Question 154:

easy

Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is:

(2020)

The limit of resolution of a telescope is given by $\Delta\theta = \frac{1.22 \lambda}{D}$. Substituting the values, $\Delta\theta = \frac{1.22 \times 600 \times 10^{-9}}{2} = 1.22 \times 300 \times 10^{-9} = 3.66 \times 10^{-7} rad$.

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Question 155:

easy

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

(2018)

Angular magnification is $M = \frac{f_o}{f_e}$, requiring a large focal length $f_o$ for the objective. Angular resolution is inversely proportional to the resolving limit $\Delta\theta = \frac{1.22\lambda}{D}$, requiring a large aperture diameter $D$ for high resolution.

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Question 156:

easy

The ratio of resolving powers of an optical microscope for two wavelengths $\lambda_1 = 4000 \AA$ and $\lambda_2 = 6000 \AA$ is

(2017-Delhi)

The resolving power of an optical microscope is inversely proportional to the wavelength of light used, $RP \propto \frac{1}{\lambda}$. Therefore, the ratio of resolving powers is $\frac{RP_1}{RP_2} = \frac{\lambda_2}{\lambda_1} = \frac{6000}{4000} = \frac{3}{2}$.

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