Tag: power of the lens

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Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of he combination, is :

  1. $+ 6.67 D$
  2. $- 6.5 D$
  3. $- 1.5 D$
  4. $+ 6.5 D$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


$f _1=+40cm$


$f _2=-25cm$

Power, $P=\dfrac{1}{f}$

The power of combination,

$P=\dfrac{1}{f _1}+\dfrac{1}{f _2}$

$P=\dfrac{100}{+40cm}+\dfrac{100}{-25cm}$

$P=-1.5D$

The correct option is C.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

If the magnitude of dispersive power of two lenses are $0.024$ and $0.036$. There focal length will be for abberation free combination.

  1. $30\ cm,\ -40\ cm$
  2. $30\ cm,\ -45\ cm$
  3. $10\ cm,\ 30\ cm$
  4. $20\ cm,\ -35\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an aberration-free combination of two thin lenses, the condition is w1/f1 + w2/f2 = 0, where w is the dispersive power and f is the focal length. Given w1=0.024 and w2=0.036, the ratio f1/f2 = -w1/w2 = -0.024/0.036 = -2/3. Option B provides f1=30 cm and f2=-45 cm, which satisfies 30/-45 = -2/3.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

In displacement method,magnification for two positions of the lens are 2 and 0.5 and the distance between the two position of the lens is 30 cm. if the focal length of lens is

  1. 15cm

  2. 20cm

  3. 25cm

  4. 30cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the displacement method, the focal length f is given by f = d(1-m^2) / (m1-m2) is incorrect; the correct formula is f = D*m / (1+m)^2 or using the displacement d and magnifications m1=2, m2=0.5. Since m1*m2 = 1, the object distance u and image distance v are swapped. The distance between positions is d = v-u = 30. With m=v/u=2, v=2u. Then 2u-u=30, so u=30, v=60. Focal length f = (u*v)/(u+v) = (30*60)/(90) = 20 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

The refractive index of the material of a double convex lens is $1.5$ and its focal lengths in $5cm$. If the radii of curvature are equal, the value of the radius of curvature is

  1. 5.0

  2. 6.5

  3. 8.0

  4. 9.5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the lens maker formula 1/f = (n-1)(1/R1 - 1/R2). For a double convex lens, R1=R and R2=-R. So 1/5 = (1.5-1)(1/R + 1/R) = 0.5 * (2/R) = 1/R. Thus R = 5 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

In a plano convex lens, the radius of curvature of the convex iens is 10 cm, if the plane side is polished , then the focal length is (Refractive index=1.5)

  1. $20.5 cm$
  2. $10 cm$
  3. $15.5 cm$
  4. $5 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a plano-convex lens with the plane side polished, it behaves like a combination of a lens and a mirror. The equivalent focal length is given by 1/f = 2/f_lens. With R = 10 cm and n = 1.5, f_lens = R/(n-1) = 20 cm, making f = 10 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

When a thin convex lens of focal length $10cm$ is kept in contact with a diverging lens, the power of the combination is found to be $-10D$. The focal length of the other lens is

  1. $-5cm$
  2. $10cm$
  3. $-25cm$
  4. $5cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power P = P1 + P2. The power of the first convex lens is 1/0.1 = +10 D. The total power is given as -10 D. Thus, -10 = 10 + P2, giving P2 = -20 D. The focal length of the second lens is f2 = 1/P2 = 1/(-20) m = -5 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

A lens made from a material of absolute refractive index $\mathrm { n } _ { 1 }$  and it is placed in a medium of absolute refractive index $\mathrm { n } _ { 2 }$   The focal length of the lens is related to $\mathrm { n } _ { 1 } \text { and } \mathrm { n } _ { 2 }$ as:

  1. $f \alpha \left( n _ { 1 } - n _ { 2 } \right)$
  2. $f \alpha \frac { 1 } { \left( n _ { 1 } - n _ { 2 } \right) }$
  3. $f \alpha \left( n _ { 1 } + n _ { 2 } \right)$
  4. $f \alpha \frac { 1 } { \left( n _ { 1 } + n _ { 2 } \right) }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to the lens maker's formula, 1/f = ((n1/n2) - 1) * (1/R1 - 1/R2). Simplifying this term shows that focal length f is inversely proportional to (n1 - n2).

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

Two thin convex lenses of focal length 10 cm  and 15 cm are separated by a distance of 10 cm. The  focal length of combination is   :-

  1. 4.2 cm

  2. 6 cm

  3. 10 cm

  4. 15 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The effective focal length of two lenses separated by a distance d is given by 1/F = 1/f1 + 1/f2 - d/(f1*f2). Substituting f1 = 10 cm, f2 = 15 cm, and d = 10 cm gives 1/F = 1/10 + 1/15 - 10/(10*15) = 3/30 + 2/30 - 2/30 = 3/30 = 1/10, so F = 10 cm.