YDSE|Special Cases|YDSE in Water|Intensity पर पुनः चर्चा|YDSE में Slit के सामने Slab रखने पर

Special Cases OF YDSE

A parrallel beam of light (lambda=5000Å) is incident at an angle alpha=30^(@) as shown in YDSE experiment. Intensity due to each slit at any point on screen in I_(0) . The distance between slits is 1mm. The intensity at central point O on the screen is KI_(0) . Find the value of K.

In YDSE experiment a uniform intensity light beam is incident on slit plane which has two slits having width ratio 9 : 4 . Find the ratio of intensities of bright and dark fringes on screen. [25]

In a YDSE with identical slits, the intensity of the central bright fringe is I_(0) . If one of the slits is covered, the intensity at the same point is

Select the correct combination of mathematical signs to sequentially replace the * signs and to balance the following equations: निम्नलिखित में से किस विकल्प के चिन्हों को * के स्थान पर रखने से दिया गया अनुक्रम संतुलित हो जाएगा ? 11*15*78*6*18*160

In YDSE of equal width slits, if intensity at the center of screen is I_(0) , then intensity at a distance of beta // 4 from the central maxima is

In YDSE, one of two slits is covered by a transparent paper which transmits only half the ligth intensity. How will the intensity of maxima and minima change ?

Consider an YDSE that has different slit width. As a result, amplitude of waves from two slits are A and 2A, respectively. If I_(0) be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is phi is