Question

37. In a double slit experiment, the two slits are
mm apart and the screen is placed 1 m away. A
monochromatic light of wavelength 500 mm is used.
What will be the width of each slit for obtaining tenth
maxima of double slit within the central maxima of
single slit pattern ?
(AIPMT 2015)
1) 0.5 mm 2) 0.02 mm 3) 0.2 mm 4) 0.1 mm
In a diffraction pattern due to a single slit of​

Answer 1

«★» ≈WAVE ≈ OPTICS ≈ YDSE

«★» ≈WAVE ≈ OPTICS ≈ YDSEExplanation:

«★» ≈WAVE ≈ OPTICS ≈ YDSEExplanation:≈ AIPMT 2015 ;

★ GIVEN ;

★ GIVEN ; » D ( Distance between slit and screen ) = 1m

★ GIVEN ; » D ( Distance between slit and screen ) = 1m » d ( Distance between 2 slits ) = 1mm = 10-³m

★ GIVEN ; » D ( Distance between slit and screen ) = 1m » d ( Distance between 2 slits ) = 1mm = 10-³m » Lambda ( wavelength ) = 500 nm

★ GIVEN ; » D ( Distance between slit and screen ) = 1m » d ( Distance between 2 slits ) = 1mm = 10-³m » Lambda ( wavelength ) = 500 nm «★» SLIT length OF " Central Maxima ( d") = ???

★ GIVEN ; » D ( Distance between slit and screen ) = 1m » d ( Distance between 2 slits ) = 1mm = 10-³m » Lambda ( wavelength ) = 500 nm «★» SLIT length OF " Central Maxima ( d") = ???»» Statement :::

★ GIVEN ; » D ( Distance between slit and screen ) = 1m » d ( Distance between 2 slits ) = 1mm = 10-³m » Lambda ( wavelength ) = 500 nm «★» SLIT length OF " Central Maxima ( d") = ???»» Statement ::: ≈ To obtain 10 Maxima's in " Central Fringe "

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1)

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)♥ NOW ; EQUATING both equations (1) and (2) ---- we get ;

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)♥ NOW ; EQUATING both equations (1) and (2) ---- we get ; » ( 10 × Lambda × D ) / d = ( 2 × Lambda × D ) / d"

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)♥ NOW ; EQUATING both equations (1) and (2) ---- we get ; » ( 10 × Lambda × D ) / d = ( 2 × Lambda × D ) / d" ≈ { Lambda and D ; gets cancelled as they are same in both equations }

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)♥ NOW ; EQUATING both equations (1) and (2) ---- we get ; » ( 10 × Lambda × D ) / d = ( 2 × Lambda × D ) / d" ≈ { Lambda and D ; gets cancelled as they are same in both equations } » 10 / d = 2 / d"

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)♥ NOW ; EQUATING both equations (1) and (2) ---- we get ; » ( 10 × Lambda × D ) / d = ( 2 × Lambda × D ) / d" ≈ { Lambda and D ; gets cancelled as they are same in both equations } » 10 / d = 2 / d" » 10 / 1 = 2 / d”

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)♥ NOW ; EQUATING both equations (1) and (2) ---- we get ; » ( 10 × Lambda × D ) / d = ( 2 × Lambda × D ) / d" ≈ { Lambda and D ; gets cancelled as they are same in both equations } » 10 / d = 2 / d" » 10 / 1 = 2 / d” »» d” = 2 / 10 = 0.2 mm ( millimeters )

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)♥ NOW ; EQUATING both equations (1) and (2) ---- we get ; » ( 10 × Lambda × D ) / d = ( 2 × Lambda × D ) / d" ≈ { Lambda and D ; gets cancelled as they are same in both equations } » 10 / d = 2 / d" » 10 / 1 = 2 / d” »» d” = 2 / 10 = 0.2 mm ( millimeters )[ANSWER] :- HENCE slit width of " Central Maxima " is 0.2 mm .

★ 10 maxima ( YDSE ) = (10 × lambda × D ) / d --------- (1) ★ Linear Width of " Central Fringe " = ( 2 × lambda × D ) / d" ---------- (2)♥ NOW ; EQUATING both equations (1) and (2) ---- we get ; » ( 10 × Lambda × D ) / d = ( 2 × Lambda × D ) / d" ≈ { Lambda and D ; gets cancelled as they are same in both equations } » 10 / d = 2 / d" » 10 / 1 = 2 / d” »» d” = 2 / 10 = 0.2 mm ( millimeters )[ANSWER] :- HENCE slit width of " Central Maxima " is 0.2 mm .♣ MILLIMETERS IS ORIGINAL LENGTH OF SLIT WIDTH ... SO UNITS WILL BE NOT IN METRES AS PER " YDSE " RULE ♣

Answer 2

Answer:

0.2 mm.

Explanation:

Angular width of a maximum in double slit pattern

$= \frac{ \beta }{D} = \frac{λ}{d}$

Angular width of central maximum in single slit pattern

$= \frac{2λ}{a} \\ given. \: 10 \times \frac{λ}{d} = \: \frac{2λ}{a} \\ = > \: a = \frac{1}{5} d = \frac{1}{5} \times 1mm \\ = 0.2mm.$

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