282399
Which one of the following statements regarding lenses is not correct?
1 A convex lens produces both real and virtual images.
2 A concave lens produces both real and virtual images.
3 A convex lens can produce images equal, greater and smaller than the size of the object.
4 A concave lens always produces images smaller than the size of the object.
Ans : (b) A concave lens produce only virtual and diminished images is not correct statement.
Explanation:
NDA (II) 2019
Ray Optics
282389
An equiconvex lens has radius of curvature ' \(R\) ', The refractive index of the material of the lens when numerical value of ' \(R\) ' and focal length ' \(f\) ' are same is
282401
An achromatic combination of lens is formed by joining
1 two convex lenses
2 two concave lens
3 one convex lens and one concave lens
4 one convex lens and one plain mirror
Explanation:
C: When a combination of lenses results in no change in focal length of the lens system for different frequencies of light, known as achromatic combination of lens.
In this case,
Power of lens is zero (0)
UPSEE 2019
Ray Optics
282390
A thin concave lens is in contact with a thin convex lens. The ratio of the magnitude of the reciprocal of focal length of concave lens and that of convex lens is \(\frac{2}{3}\). The focal length of the combined lens is \(30 \mathrm{~cm}\). Focal length of the concave lens is
1 \(-15 \mathrm{~cm}\)
2 red light
3 yellow light
4 blue light
#[examname: MHT-CET 2020]#
Explanation:
A: Let,
Focal length of convex lens \(=f_1\)
Focal length of concave lens \(=-f_2\)
Given,
\(\frac{1 / \mathrm{f}_2}{1 / \mathrm{f}_1}=\frac{2}{3}=\frac{\mathrm{P}_2}{\mathrm{P}_1}\) or \(\quad \frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{2}{3}\)
Combined focal length, \(\mathrm{F}=30 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{30}=\frac{3}{2 \mathrm{f}_2}-\frac{1}{\mathrm{f}_2} \\
\frac{1}{30}=\frac{(3-2)}{2 \mathrm{f}_2} \\
2 \mathrm{f}_2=30 \\
\mathrm{f}_2=15 \mathrm{~cm}
\end{aligned}\)
(by magnitude)
Hence, \(\mathrm{f}_2=-15 \mathrm{~cm}\) (because of concave lens) 496. Focal length of a convex lens will be minimum for(a) violet light
(b.) red light
(c.) yellow light
(d.) blue light
#[examname: MHT-CET 2020]#
Ans: a
Exp: A: Wavelength for violet light is minimum, so it bends maximum.
Hence, focal length of violet light is minimum among following.
282399
Which one of the following statements regarding lenses is not correct?
1 A convex lens produces both real and virtual images.
2 A concave lens produces both real and virtual images.
3 A convex lens can produce images equal, greater and smaller than the size of the object.
4 A concave lens always produces images smaller than the size of the object.
Ans : (b) A concave lens produce only virtual and diminished images is not correct statement.
Explanation:
NDA (II) 2019
Ray Optics
282389
An equiconvex lens has radius of curvature ' \(R\) ', The refractive index of the material of the lens when numerical value of ' \(R\) ' and focal length ' \(f\) ' are same is
282401
An achromatic combination of lens is formed by joining
1 two convex lenses
2 two concave lens
3 one convex lens and one concave lens
4 one convex lens and one plain mirror
Explanation:
C: When a combination of lenses results in no change in focal length of the lens system for different frequencies of light, known as achromatic combination of lens.
In this case,
Power of lens is zero (0)
UPSEE 2019
Ray Optics
282390
A thin concave lens is in contact with a thin convex lens. The ratio of the magnitude of the reciprocal of focal length of concave lens and that of convex lens is \(\frac{2}{3}\). The focal length of the combined lens is \(30 \mathrm{~cm}\). Focal length of the concave lens is
1 \(-15 \mathrm{~cm}\)
2 red light
3 yellow light
4 blue light
#[examname: MHT-CET 2020]#
Explanation:
A: Let,
Focal length of convex lens \(=f_1\)
Focal length of concave lens \(=-f_2\)
Given,
\(\frac{1 / \mathrm{f}_2}{1 / \mathrm{f}_1}=\frac{2}{3}=\frac{\mathrm{P}_2}{\mathrm{P}_1}\) or \(\quad \frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{2}{3}\)
Combined focal length, \(\mathrm{F}=30 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{30}=\frac{3}{2 \mathrm{f}_2}-\frac{1}{\mathrm{f}_2} \\
\frac{1}{30}=\frac{(3-2)}{2 \mathrm{f}_2} \\
2 \mathrm{f}_2=30 \\
\mathrm{f}_2=15 \mathrm{~cm}
\end{aligned}\)
(by magnitude)
Hence, \(\mathrm{f}_2=-15 \mathrm{~cm}\) (because of concave lens) 496. Focal length of a convex lens will be minimum for(a) violet light
(b.) red light
(c.) yellow light
(d.) blue light
#[examname: MHT-CET 2020]#
Ans: a
Exp: A: Wavelength for violet light is minimum, so it bends maximum.
Hence, focal length of violet light is minimum among following.
282399
Which one of the following statements regarding lenses is not correct?
1 A convex lens produces both real and virtual images.
2 A concave lens produces both real and virtual images.
3 A convex lens can produce images equal, greater and smaller than the size of the object.
4 A concave lens always produces images smaller than the size of the object.
Ans : (b) A concave lens produce only virtual and diminished images is not correct statement.
Explanation:
NDA (II) 2019
Ray Optics
282389
An equiconvex lens has radius of curvature ' \(R\) ', The refractive index of the material of the lens when numerical value of ' \(R\) ' and focal length ' \(f\) ' are same is
282401
An achromatic combination of lens is formed by joining
1 two convex lenses
2 two concave lens
3 one convex lens and one concave lens
4 one convex lens and one plain mirror
Explanation:
C: When a combination of lenses results in no change in focal length of the lens system for different frequencies of light, known as achromatic combination of lens.
In this case,
Power of lens is zero (0)
UPSEE 2019
Ray Optics
282390
A thin concave lens is in contact with a thin convex lens. The ratio of the magnitude of the reciprocal of focal length of concave lens and that of convex lens is \(\frac{2}{3}\). The focal length of the combined lens is \(30 \mathrm{~cm}\). Focal length of the concave lens is
1 \(-15 \mathrm{~cm}\)
2 red light
3 yellow light
4 blue light
#[examname: MHT-CET 2020]#
Explanation:
A: Let,
Focal length of convex lens \(=f_1\)
Focal length of concave lens \(=-f_2\)
Given,
\(\frac{1 / \mathrm{f}_2}{1 / \mathrm{f}_1}=\frac{2}{3}=\frac{\mathrm{P}_2}{\mathrm{P}_1}\) or \(\quad \frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{2}{3}\)
Combined focal length, \(\mathrm{F}=30 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{30}=\frac{3}{2 \mathrm{f}_2}-\frac{1}{\mathrm{f}_2} \\
\frac{1}{30}=\frac{(3-2)}{2 \mathrm{f}_2} \\
2 \mathrm{f}_2=30 \\
\mathrm{f}_2=15 \mathrm{~cm}
\end{aligned}\)
(by magnitude)
Hence, \(\mathrm{f}_2=-15 \mathrm{~cm}\) (because of concave lens) 496. Focal length of a convex lens will be minimum for(a) violet light
(b.) red light
(c.) yellow light
(d.) blue light
#[examname: MHT-CET 2020]#
Ans: a
Exp: A: Wavelength for violet light is minimum, so it bends maximum.
Hence, focal length of violet light is minimum among following.
282399
Which one of the following statements regarding lenses is not correct?
1 A convex lens produces both real and virtual images.
2 A concave lens produces both real and virtual images.
3 A convex lens can produce images equal, greater and smaller than the size of the object.
4 A concave lens always produces images smaller than the size of the object.
Ans : (b) A concave lens produce only virtual and diminished images is not correct statement.
Explanation:
NDA (II) 2019
Ray Optics
282389
An equiconvex lens has radius of curvature ' \(R\) ', The refractive index of the material of the lens when numerical value of ' \(R\) ' and focal length ' \(f\) ' are same is
282401
An achromatic combination of lens is formed by joining
1 two convex lenses
2 two concave lens
3 one convex lens and one concave lens
4 one convex lens and one plain mirror
Explanation:
C: When a combination of lenses results in no change in focal length of the lens system for different frequencies of light, known as achromatic combination of lens.
In this case,
Power of lens is zero (0)
UPSEE 2019
Ray Optics
282390
A thin concave lens is in contact with a thin convex lens. The ratio of the magnitude of the reciprocal of focal length of concave lens and that of convex lens is \(\frac{2}{3}\). The focal length of the combined lens is \(30 \mathrm{~cm}\). Focal length of the concave lens is
1 \(-15 \mathrm{~cm}\)
2 red light
3 yellow light
4 blue light
#[examname: MHT-CET 2020]#
Explanation:
A: Let,
Focal length of convex lens \(=f_1\)
Focal length of concave lens \(=-f_2\)
Given,
\(\frac{1 / \mathrm{f}_2}{1 / \mathrm{f}_1}=\frac{2}{3}=\frac{\mathrm{P}_2}{\mathrm{P}_1}\) or \(\quad \frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{2}{3}\)
Combined focal length, \(\mathrm{F}=30 \mathrm{~cm}\)
\(\begin{aligned}
\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\
\frac{1}{30}=\frac{3}{2 \mathrm{f}_2}-\frac{1}{\mathrm{f}_2} \\
\frac{1}{30}=\frac{(3-2)}{2 \mathrm{f}_2} \\
2 \mathrm{f}_2=30 \\
\mathrm{f}_2=15 \mathrm{~cm}
\end{aligned}\)
(by magnitude)
Hence, \(\mathrm{f}_2=-15 \mathrm{~cm}\) (because of concave lens) 496. Focal length of a convex lens will be minimum for(a) violet light
(b.) red light
(c.) yellow light
(d.) blue light
#[examname: MHT-CET 2020]#
Ans: a
Exp: A: Wavelength for violet light is minimum, so it bends maximum.
Hence, focal length of violet light is minimum among following.
