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3 years ago

6

What type of specialized cell in the eye is used for detecting low levels of light?

2 answers:

Hunter-Best [27]3 years ago

8 0

Answer:

There are about six to seven million cones in a human eye and are most concentrated towards the macula. Cones are less sensitive to light than the rod cells in the retina (which support vision at low light levels), but allow the perception of color.

pashok25 [27]3 years ago

8 0

Answer:

Rods

Explanation:

There are about six to seven million cones in a human eye and are most concentrated towards the macula. Cones are less sensitive to light than the rod cells in the retina (which support vision at low light levels), but allow the perception of color.

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VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers $d=1.6\ m$

Distance of person from one speaker $x_1=3\ m$

Distance of person from second speaker $x_2=3.5\ m$

Path difference between the waves is given by

$x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}$

for destructive interference m=0 I.e.

$x_2-x_1=\frac{\lambda }{2}$

$3.5-3=\frac{\lambda }{2}$

$\lambda =0.5\times 2$

$\lambda =1\ m$

frequency is given by

$f=\frac{v}{\lambda }$

where $v=velocity\ of\ sound\ (v=343\ m/s)$

$f=\frac{343}{1}=343\ Hz$

For next frequency which will cause destructive interference is

i.e. $m=1$ and $m=2$

$3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda$

$\lambda =\frac{1}{3}\ m$

frequency corresponding to this is

$f_2=\frac{343}{\frac{1}{3}}=1029\ Hz$

for $m=2$

$3.5-3=\frac{5}{2}\cdot \lambda$

$\lambda =\frac{1}{5}\ m$

Frequency corresponding to this wavelength

$f_3=\frac{343}{\frac{1}{5}}$

$f_3=1715\ Hz$

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