
The luminous intensity of a light source is 500cd. The illuminance of a surface distant 10m from it, will be if light falls normally on it
A. 5 lux
B. 10 lux
C. 20 lux
D. 40 lux
Answer
159.3k+ views
Hint: In this question, we are given the luminous intensity of a light source and the distance of the light source from the screen. We know that the illuminance of a light source is defined as the luminous intensity falling on per unit square area. We just need to substitute the given values in this equation.
Complete Step by step solution:
Luminous intensity is the amount of visible light emitted by any light source in unit time per unit solid angle. It is one of the seven fundamental physical quantities in physics. The SI unit for luminous intensity is Candela, cd.
Illuminance is the total luminous flux incident on any surface per unit area of the surface. It is measured in lux.
Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], where I is the luminous intensity and R is the distance between the surface and the light source
Given that \[I = 500cd\,and\,R = 10m\]
By substituting the values, we get that
\[ \Rightarrow E = \dfrac{{500}}{{{{10}^2}}} = 5lux\]
Note: We know that Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], is different from Luminous Intensity that is I. Illuminance is different from brightness as illuminance is the measure of intensity of light falling onto a surface, while brightness is the visual perceptions of light.
Complete Step by step solution:
Luminous intensity is the amount of visible light emitted by any light source in unit time per unit solid angle. It is one of the seven fundamental physical quantities in physics. The SI unit for luminous intensity is Candela, cd.
Illuminance is the total luminous flux incident on any surface per unit area of the surface. It is measured in lux.
Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], where I is the luminous intensity and R is the distance between the surface and the light source
Given that \[I = 500cd\,and\,R = 10m\]
By substituting the values, we get that
\[ \Rightarrow E = \dfrac{{500}}{{{{10}^2}}} = 5lux\]
Note: We know that Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], is different from Luminous Intensity that is I. Illuminance is different from brightness as illuminance is the measure of intensity of light falling onto a surface, while brightness is the visual perceptions of light.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Charging and Discharging of Capacitor

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
