A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. The central bright fringe becomes narrower. The effect cannot be determined unless the distance between the slit and the screen is known. The central bright fringe becomes wider. The central bright fringe remains the same size.

Explanation:

General guidance

Concepts and reason

The concept used to solve this problem is slit width condition for maximum diffraction in case of single slit diffraction experiment.

Initially, use the condition for diffraction maximum in the case of single slit diffraction to find the inapplicable given options.

Finally, use the condition for diffraction maximum in the case of single slit diffraction to find the applicable given options.

Fundamentals

The condition for diffraction maximum in the case of single slit diffraction is as follows:

sin Θ=λ/α

Here, the angle situated in the first dark fringe on each side of the central bright fringe isΘ , slit width is α, and the wavelength is λ .

The incorrect options are as follows:

• The central bright fringe remains in same size.

The width of the central bright fringe is inversely proportional to the slit width. Therefore, the central fringe cannot remain in the same size. Hence, it is incorrect.

• The effect cannot be determined unless the distance between the slit and screen is known.

Without knowing the distance between the slit and screen, the effect can be experienced. Therefore, this option is incorrect.

• The central bright fringe becomes narrower.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becoming narrower is incorrect.The central bright fringe width is directly proportional to the slit width.

If the width of the slit increases, then the central bright fringe width also increases.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becomes wider when the width of the single slit is reduced.

The condition for diffraction maximum is as follows:

sin  Θ=λ/α

The slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

Therefore,

• The central bright fringe becomes wider is correct.

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.  

Slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

The central bright fringe width is directly proportional to the angle of the first dark fringe on either side of the central bright fringe.

If the central bright fringe becomes wider, then the angle of the first dark fringe on either side of the central bright fringe will be larger.

Answer

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.