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Question

White light is incident normally on a thin film which has n = 1.5 and a thickness of 5000 Å. For what wavelengths in the visible spectrum (4000 – 7000 Å) will the intensity of the reflected light be a maximum?

5000 Å, 6000 Å
5260 Å, 6000 Å
4285 Å, 6000 Å
4500 Å, 6500 Å

The correct answer is: 4285 Å, 6000 Å

Constructive interference occurs for
OPD = 2nt = k + $fraction numerator lambda over denominator 2 end fraction$ ,
where t is the thickness of the film, k an integer. Hence for maximum reflection the wavelengths are
 = $fraction numerator 4 n t over denominator 2 k plus 1 end fraction$ = $open curly brackets table row cell 6000 Å f o r k equals 2 end cell row cell 4285 Å f o r k equals 3. end cell end table close$

The sum of 730 + 450 using front – end estimation is______.

Front-end estimation means rounding off a number to the nearest place value of the leading digit of the number.

The sum of 730 + 450 using front – end estimation is______.

Maths-General

Front-end estimation means rounding off a number to the nearest place value of the leading digit of the number.

General

Maths-Video

Round the number 5432 to the nearest hundred.

For rounding off to the nearest hundred, we can also look at the number formed by the tens place digit and one place digit. If the number formed is less than 50,we round down and if the number is 50 or more we round up.

Round the number 5432 to the nearest hundred.

Maths-General

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GeneralVideo

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What is the value of 20/100?

MathematicsGeneral

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Identify structures A to D –

BiologyGeneral

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botany-

An ellipse slides between two perpendicular straight lines $x equals 0 blank a n d blank Y equals 0$ then, locus of its foci is

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If there are exactly two points on the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ whose distance from its centre is same and is equal to $square root of fraction numerator a to the power of 2 end exponent plus 2 b to the power of 2 end exponent over denominator 2 end fraction end root$ then eccentricity of the ellipse is

botany-General

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botany-

Tangent at any point of ellipse $9 x to the power of 2 end exponent plus 16 y to the power of 2 end exponent minus 144 equals 0$ is drawn. Eccentric
angle of ‘ P’ is $theta equals fraction numerator 1 over denominator 2 end fraction s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator 7 end fraction close parentheses$ If is the foot of perpendicular from centre $blank ´ O blank ´$ to this tangent then $angle P O N$ is

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If the curve $x to the power of 2 end exponent plus 3 y to the power of 2 end exponent equals 9$ subtends as obtuse angle at the point $left parenthesis 2 alpha comma alpha right parenthesis left parenthesis left parenthesis alpha element of text int end text text eger end text text end text right parenthesis$ , $blank left parenthesis alpha element of text int eger end text right parenthesis$ then a possible value of is

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Let $Q equals left parenthesis 3 comma square root of 5 right parenthesis comma R equals left parenthesis 7 , 3 square root of 5 right parenthesis$ . A point P in the XY-plane varies in such a way that perimeter of $capital delta P Q R$ is 16. Then the maximum area of $capital delta P Q R$ is

botany-General

General

botany-

$P comma Q blank$ are points on the ellipse $fraction numerator x to the power of 2 end exponent over denominator 25 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 16 end fraction equals 1$ such that $P Q$ is a chord through the point $R left parenthesis 3 , 0 right parenthesis text . end text$ If $vertical line P R vertical line equals 2$ then length of chord $P Q$ is

botany-General

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Can’t find what you’re looking for?

White light is incident normally on a thin film which has n = 1.5 and a thickness of 5000 Å. For what wavelengths in the visible spectrum (4000 – 7000 Å) will the intensity of the reflected light be a maximum?

5000 Å, 6000 Å 5260 Å, 6000 Å 4285 Å, 6000 Å 4500 Å, 6500 Å

The correct answer is: 4285 Å, 6000 Å

Constructive interference occurs forOPD = 2nt = k +,where t is the thickness of the film, k an integer. Hence for maximum reflection the wavelengths are ==

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Round the number 5432 to the nearest hundred.