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The focal lengths of a converging lens m...

The focal lengths of a converging lens measured for violet, green and red colours of `fv'f_(G)'f_(R)` respectively. We will find
(A). `f_(R)gtf_(G)gtf_(V)` (B). `f_(v)ltf_(R)ltf_(G)`
(C). `f_(V)gtf_(R)gtf_(G)` (D). `f_(V)=f_(R)=f_(G)`

A

`f_(R)gtf_(G)gtf_(V)`

B

`f_(v)ltf_(R)ltf_(G)`

C

`f_(V)gtf_(R)gtf_(G)`

D

`f_(V)=f_(R)=f_(G)`

Text Solution

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The correct Answer is:
To solve the problem regarding the focal lengths of a converging lens for different colors of light (violet, green, and red), we will follow these steps: ### Step 1: Understand the relationship between wavelength and refractive index The refractive index (μ) of a lens material is inversely proportional to the wavelength (λ) of light. As the wavelength increases, the refractive index decreases. This can be expressed as: \[ \mu \propto \frac{1}{\lambda} \] ### Step 2: Apply the lens maker's formula According to the lens maker's formula, the focal length (f) of a lens is related to its refractive index and the radii of curvature of its surfaces. The formula can be simplified (assuming the surrounding medium has a refractive index of 1) as: \[ \frac{1}{f} \propto \mu - 1 \] This implies that: \[ f \propto \frac{1}{\mu} \] ### Step 3: Combine the relationships From the above two relationships, we can combine them to show that the focal length of the lens is directly proportional to the wavelength: \[ f \propto \lambda \] This means that as the wavelength increases, the focal length also increases. ### Step 4: Compare the wavelengths of the colors In the visible spectrum, the order of wavelengths from shortest to longest is: - Violet (λ is the shortest) - Green (λ is medium) - Red (λ is the longest) Since the focal length is directly proportional to the wavelength, we can conclude: - \( f_R > f_G > f_V \) ### Step 5: Select the correct option Based on our findings, the correct relationship among the focal lengths for red, green, and violet light is: \[ f_R > f_G > f_V \] Thus, the correct option is: **(A) \( f_R > f_G > f_V \)** ---
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The image of a white object in with light formed by a lens is usually colored and blurred. This defect of image is called chromatic aberration and arises due to the fact that focal length of a lens is different for different colours. As R . I . mu of lens is maximum for violet while minimum for red, violet is focused nearest to the lens while red farthest from it as shown in figure. As a result of this, in case of convergent lens if a screen is placed at F_(v) center of the image will be violet and focused while sides are red and blurred. While at F_(R) , reverse is the case, i.e ., center will be red and focused while sides violet and blurred. The differece between f_(v) and f_(R) is a measure of the longitudinal chromatic aberration (L.C.A),i.e., L.C.A.=f_(R)-f_(v)=-df with df=f_(v)-f_(R) ........... (1) However, as for a single lens, (1)/(f)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))] ............. (2) rArr -(df)/(f^(2))=dmu[(1)/(R_(1))-(1)/(R_(2))] ............... (3) Dividing E1n. (3) by (2) : -(df)/(f)=(dmu)/((mu-1))=omega, [omega=(dmu)/((mu-1))] "dispersive power" , .........(4) And hence, from Eqns. (1) and (4) , L.C.A.=-df=omegaf Now, as for a single lens neither f nor omega zero, we cannot have a single lens free from chromatic aberration. Condition of Achromatism : In case of two thin lenses in contact (1)/(F)=(1)/(F_(1))+(1)/(F_(2)) i.c,. -(dF)/(F^(2))=(df_(1))/(f_(1)^(2))-(df_(2))/(f_(2)^(2)) The combination will be free from chromatic aberration if dF=0 i.e., (df_(1))/(f_(1)^(2))+(df_(2))/(f_(2)^(2))=0 which with the help of Eqn. (4) reduces to (omega_(1)f_(1))/(f_(1)^(2))+(omega_(2)f_(2))/(f_(2)^(2))=0 , i.e., (omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0 ........(5) This condition is called condition of achromatism (for two thin lenses in contact ) and the lens combination which satisfies this condition is called achromatic lems, from this condition, i.e., form Eqn. (5) it is clear the in case of achromatic doublet : Since, if omega_(1)=omega_(2), (1)/(f_(1))+(1)/(f_(2))=0 i.e., (1)/(F)=0 or F=infty i.e., combination will not behave as a lens, but as a plane glass plate. (2) As omega_(1) and omega_(2) are positive quantities, for equation (5) to hold, f_(1) and f_(2) must be of opposite nature, i.e., if one of the lenses is converging the other must be diverging. (3) If the achromatic combination is convergent, f_(C)ltf_(D) and as (f_(C))/(f_(d))=(omega_(C))/(omega_(D)), omega_(C)ltomega_(d) i.e., in a convergent achromatic doublet, convex lens has lesses focal legth and dispersive power than the divergent one. A combination is made of two lenses of focal lengths f and f' in contact , the dispersive powers of the materials of the lenses are omega and omega' . The combination is achromatic when :

The image of a white object in with light formed by a lens is usually colored and blurred. This defect of image is called chromatic aberration and arises due to the fact that focal length of a lens is different for different colours. As R . I . mu of lens is maximum for violet while minimum for red, violet is focused nearest to the lens while red farthest from it as shown in figure. As a result of this, in case of convergent lens if a screen is placed at F_(v) center of the image will be violet and focused while sides are red and blurred. While at F_(R) , reverse is the case, i.e ., center will be red and focused while sides violet and blurred. The differece between f_(v) and f_(R) is a measure of the longitudinal chromatic aberration (L.C.A),i.e., L.C.A.=f_(R)-f_(v)=-df with df=f_(v)-f_(R) ........... (1) However, as for a single lens, (1)/(f)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))] ............. (2) rArr -(df)/(f^(2))=dmu[(1)/(R_(1))-(1)/(R_(2))] ............... (3) Dividing E1n. (3) by (2) : -(df)/(f)=(dmu)/((mu-1))=omega, [omega=(dmu)/((mu-1))] "dispersive power" , .........(4) And hence, from Eqns. (1) and (4) , L.C.A.=-df=omegaf Now, as for a single lens neither f nor omega zero, we cannot have a single lens free from chromatic aberration. Condition of Achromatism : In case of two thin lenses in contact (1)/(F)=(1)/(F_(1))+(1)/(F_(2)) i.c,. -(dF)/(F^(2))=(df_(1))/(f_(1)^(2))-(df_(2))/(f_(2)^(2)) The combination will be free from chromatic aberration if dF=0 i.e., (df_(1))/(f_(1)^(2))+(df_(2))/(f_(2)^(2))=0 which with the help of Eqn. (4) reduces to (omega_(1)f_(1))/(f_(1)^(2))+(omega_(2)f_(2))/(f_(2)^(2))=0 , i.e., (omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0 ........(5) This condition is called condition of achromatism (for two thin lenses in contact ) and the lens combination which satisfies this condition is called achromatic lems, from this condition, i.e., form Eqn. (5) it is clear the in case of achromatic doublet : Since, if omega_(1)=omega_(2), (1)/(f_(1))+(1)/(f_(2))=0 i.e., (1)/(F)=0 or F=infty i.e., combination will not behave as a lens, but as a plane glass plate. (2) As omega_(1) and omega_(2) are positive quantities, for equation (5) to hold, f_(1) and f_(2) must be of opposite nature, i.e., if one of the lenses is converging the other must be diverging. (3) If the achromatic combination is convergent, f_(C)ltf_(D) and as (f_(C))/(f_(d))=(omega_(C))/(omega_(D)), omega_(C)ltomega_(d) i.e., in a convergent achromatic doublet, convex lens has lesses focal legth and dispersive power than the divergent one. Chromatic aberration of a lens can be corrected by :

The image of a white object in with light formed by a lens is usually colored and blurred. This defect of image is called chromatic aberration and arises due to the fact that focal length of a lens is different for different colours. As R . I . mu of lens is maximum for violet while minimum for red, violet is focused nearest to the lens while red farthest from it as shown in figure. As a result of this, in case of convergent lens if a screen is placed at F_(v) center of the image will be violet and focused while sides are red and blurred. While at F_(R) , reverse is the case, i.e ., center will be red and focused while sides violet and blurred. The differece between f_(v) and f_(R) is a measure of the longitudinal chromatic aberration (L.C.A),i.e., L.C.A.=f_(R)-f_(v)=-df with df=f_(v)-f_(R) ........... (1) However, as for a single lens, (1)/(f)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))] ............. (2) rArr -(df)/(f^(2))=dmu[(1)/(R_(1))-(1)/(R_(2))] ............... (3) Dividing E1n. (3) by (2) : -(df)/(f)=(dmu)/((mu-1))=omega, [omega=(dmu)/((mu-1))] "dispersive power" , .........(4) And hence, from Eqns. (1) and (4) , L.C.A.=-df=omegaf Now, as for a single lens neither f nor omega zero, we cannot have a single lens free from chromatic aberration. Condition of Achromatism : In case of two thin lenses in contact (1)/(F)=(1)/(F_(1))+(1)/(F_(2)) i.c,. -(dF)/(F^(2))=(df_(1))/(f_(1)^(2))-(df_(2))/(f_(2)^(2)) The combination will be free from chromatic aberration if dF=0 i.e., (df_(1))/(f_(1)^(2))+(df_(2))/(f_(2)^(2))=0 which with the help of Eqn. (4) reduces to (omega_(1)f_(1))/(f_(1)^(2))+(omega_(2)f_(2))/(f_(2)^(2))=0 , i.e., (omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0 ........(5) This condition is called condition of achromatism (for two thin lenses in contact ) and the lens combination which satisfies this condition is called achromatic lems, from this condition, i.e., form Eqn. (5) it is clear the in case of achromatic doublet : Since, if omega_(1)=omega_(2), (1)/(f_(1))+(1)/(f_(2))=0 i.e., (1)/(F)=0 or F=infty i.e., combination will not behave as a lens, but as a plane glass plate. (2) As omega_(1) and omega_(2) are positive quantities, for equation (5) to hold, f_(1) and f_(2) must be of opposite nature, i.e., if one of the lenses is converging the other must be diverging. (3) If the achromatic combination is convergent, f_(C)ltf_(D) and as (f_(C))/(f_(d))=(omega_(C))/(omega_(D)), omega_(C)ltomega_(d) i.e., in a convergent achromatic doublet, convex lens has lesses focal legth and dispersive power than the divergent one. The dispersive power of crown and fint glasses are 0.02 and 0.04 respectively. An achromtic converging lens of focal length 40 cm is made by keeping two lenses, one of crown glass and the other of flint glass, in contact with each other. The focal lengths of the two lenses are :

The image of a white object in with light formed by a lens is usually colored and blurred. This defect of image is called chromatic aberration and arises due to the fact that focal length of a lens is different for different colours. As R . I . mu of lens is maximum for violet while minimum for red, violet is focused nearest to the lens while red farthest from it as shown in figure. As a result of this, in case of convergent lens if a screen is placed at F_(v) center of the image will be violet and focused while sides are red and blurred. While at F_(R) , reverse is the case, i.e ., center will be red and focused while sides violet and blurred. The differece between f_(v) and f_(R) is a measure of the longitudinal chromatic aberration (L.C.A),i.e., L.C.A.=f_(R)-f_(v)=-df with df=f_(v)-f_(R) ........... (1) However, as for a single lens, (1)/(f)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))] ............. (2) rArr -(df)/(f^(2))=dmu[(1)/(R_(1))-(1)/(R_(2))] ............... (3) Dividing E1n. (3) by (2) : -(df)/(f)=(dmu)/((mu-1))=omega, [omega=(dmu)/((mu-1))] "dispersive power" , .........(4) And hence, from Eqns. (1) and (4) , L.C.A.=-df=omegaf Now, as for a single lens neither f nor omega zero, we cannot have a single lens free from chromatic aberration. Condition of Achromatism : In case of two thin lenses in contact (1)/(F)=(1)/(F_(1))+(1)/(F_(2)) i.c,. -(dF)/(F^(2))=(df_(1))/(f_(1)^(2))-(df_(2))/(f_(2)^(2)) The combination will be free from chromatic aberration if dF=0 i.e., (df_(1))/(f_(1)^(2))+(df_(2))/(f_(2)^(2))=0 which with the help of Eqn. (4) reduces to (omega_(1)f_(1))/(f_(1)^(2))+(omega_(2)f_(2))/(f_(2)^(2))=0 , i.e., (omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0 ........(5) This condition is called condition of achromatism (for two thin lenses in contact ) and the lens combination which satisfies this condition is called achromatic lems, from this condition, i.e., form Eqn. (5) it is clear the in case of achromatic doublet : Since, if omega_(1)=omega_(2), (1)/(f_(1))+(1)/(f_(2))=0 i.e., (1)/(F)=0 or F=infty i.e., combination will not behave as a lens, but as a plane glass plate. (2) As omega_(1) and omega_(2) are positive quantities, for equation (5) to hold, f_(1) and f_(2) must be of opposite nature, i.e., if one of the lenses is converging the other must be diverging. (3) If the achromatic combination is convergent, f_(C)ltf_(D) and as (f_(C))/(f_(d))=(omega_(C))/(omega_(D)), omega_(C)ltomega_(d) i.e., in a convergent achromatic doublet, convex lens has lesses focal legth and dispersive power than the divergent one. Chromatic aberration in the formation of image by a lens arises because :

If f,g,\ h are three functions defined from R\ to\ R as follows: Find the range of f(x)=x^2