Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
Answer:
d d d d dd d d d d d dd d d d d dd d
Hi!
The correct answer would be: the width of I-bands
The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).
<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>
<u>When a sacromere contracts, a series of changes take place which include:</u>
<em>- Shortening of I band, and consequently the H zone</em>
<em>- The A line remains unchanged</em>
<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>
The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.
Hope this helps!
The correct answer is going to be <span>C, because in the nucleus of an atom there are protons and electrons; which can't move, and are surrounded by electrons on the electrical cloud</span>