101.
In $$YDSE$$ distance between the $${S_1}$$ and $${S_2}$$ is $$d.$$ $${P_1}$$ and $${P_2}$$ are two points equidistance from $$O$$ at an angular position $$\beta $$ as shown. A parallel beam of monochromatic light is incident at an angle $$\alpha $$ on the slits. Then the ratio of path difference at $${P_1}$$ and $${P_2}$$ is:
A
$$\cot \frac{{\alpha - \beta }}{2}\cot \frac{{\alpha + \beta }}{2}$$
B
$$\tan \frac{{\alpha + \beta }}{2}\cot \frac{{\alpha - \beta }}{2}$$
C
$$\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}$$
D
$$\tan \frac{{\alpha - \beta }}{2}\cot \frac{{\alpha + \beta }}{2}$$
102.
In $$YSDE,$$ both slits are covered by transparent slab. Upper slit is covered by slab of $$R.I.$$ 1.5 and thickness $$t$$ and lower is covered by $$R.I.$$ $$\frac{4}{3}$$ and thickness $$2t,$$ then central maxima
A
shifts in $$+ve$$ $$y$$ -axis direction
B
shifts in $$-ve$$ $$y$$ -axis direction
C
remains at same position
D
may shift in upward or downward depending upon wavelength of light
103.
In a Young's double slit experiment, slits are separated by $$0.5\,mm,$$ and the screen is placed $$150\,cm$$ away. A beam of light consisting of two wavelengths, $$650\,nm$$ and $$520\,nm,$$ is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:
104.
A point source of light $$B$$ is placed at a distance $$L$$ in front of the centre of a mirror of width $$'d’$$ hung vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance $$2\,L$$ from it as shown in fig. The greatest distance over which he can see the image of the light source in the mirror is
From the ray diagram.
$$\eqalign{
& {\text{In }}\Delta \,ANM\,\,{\text{and }}\Delta \,ADB \cr
& \angle \,ADB = \angle \,ANM = {90^ \circ } \cr
& \angle \,MAN = \angle \,BAN\,\,\left( {{\text{laws of reflection}}} \right) \cr
& {\text{Also, }}\angle \,BAN = \angle \,ABD \cr
& \Rightarrow \,\,\angle \,MAN = \angle \,ABD \cr} $$
∴ $$\Delta \,ANM$$ is similar to $$\Delta \,ADB$$
$$\therefore \,\,\frac{x}{{2\,L}} = \frac{{\frac{d}{2}}}{L}\,\,{\text{or }}x = d$$
So, required distance $$= d + d + d = 3\,d.$$
105.
Yellow light is used in a single slit diffraction experiment with slit width of $$0.6\,mm.$$ If yellow light is replaced by $$X$$ - rays, then the observed pattern will reveal,
For diffraction pattern to be observed, the dimension of slit should be comparable to the wave length of rays. The wavelength of $$X$$ - rays $$\left( {1 - 100\,\mathop {\text{A}}\limits^ \circ } \right)$$ is less than $$0.6\,mm.$$
106.
In the ideal double - slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wave - lenght $$\lambda $$ ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is
107.
A radio transmitting station operating at a frequency of $$120\,MHz$$ has two identical antennas that radiate in phase. Antenna $$B$$ is $$9\,m$$ to the right of antenna $$A.$$ Consider point $$P$$ at a horizontal distance $$x$$ to the right of antenna $$A$$ as shown in Fig. The value of $$x$$ and order for which the constructive interference will occur at point $$P$$ is
For the phenomenon of interference we require two sources of light of same frequency and having a definite phase relationship (a phase relationship that does not change with time)
109.
In the figure shown in a $$YDSE,$$ a parallel beam of light is incident on the slits from a medium of refractive index $${n_1}.$$ The wavelength of light in this medium is $${\lambda _1}.$$ A transparent slab of thickness $$t$$ and refractive index is put infront of one slit. The medium between the screen and the plane of the slits is $${n_2}.$$ The phase difference between the light waves reaching point $$O$$ (symmetrical, relative to the slit) is
A
$$\frac{{2\pi }}{{{n_1}{\lambda _1}}}\left( {{n_3} - {n_2}} \right)t$$
B
$$\frac{{2\pi }}{{{\lambda _1}}}\left( {{n_3} - {n_2}} \right)t$$
C
$$\frac{{2\pi {n_1}}}{{{n_2}{\lambda _1}}}\left( {\frac{{{n_3}}}{{{n_2}}} - 1} \right)t$$
D
$$\frac{{2\pi {n_1}}}{{{\lambda _1}}}\left( {{n_3} - {n_2}} \right)t$$
110.
Figure shows two coherent sources $${S_1} - {S_2}$$ vibrating in same phase. $$AB$$ is an irregular wire lying at a far distance from the sources $${S_1}$$ and $${S_2}.$$ Let $$\frac{\lambda }{d} = {10^{ - 3}}.\angle BOA = {0.12^ \circ }.$$ How many bright spots will be seen on the wire, including points $$A$$ and $$B.$$
Angular width $$ = \frac{\lambda }{d} = {10^{ - 3}}\left( {{\text{given}}} \right)$$
$$\therefore $$ No. of fringes within $${0.12^ \circ }$$ will be $$n = \frac{{0.12 \times 2\pi }}{{360 \times {{10}^{ - 3}}}} \cong \left[ {2.09} \right]$$
$$\therefore $$ The number of bright spots will be two