Power of the lens - class-XII

For the lenses placed coaxially, 

The power of the combination is P = P1 + P2 = 1/f1 + 1/f2. Here, 1/20 + 1/(-20) = 0. Since P = 1/f, f = 1/0 = infinity.

The apparent shift due to a glass slab is given by t(1 - 1/n). Here, 3(1 - 1/(3/2)) = 3(1 - 2/3) = 3(1/3) = 1 cm. The mark appears 1 cm closer to the surface. Since the person is 2 cm above the slab (total distance 5 cm from the mark), the apparent distance is 5 - 1 = 4 cm.

Total power, $P _{net} = P _1 + P _2 + P _3$

Abeam of a parallel rays is brought to a focus by convex lens. Now, when thin concave lens of equal focal length is joined to first lens, then combined focal length be

$P _{original} = 4D$

$f$ will be less than $f _1$

If resulting focus is $f$ then $ \dfrac{1}{f} = \dfrac{1}{f _1} + \dfrac{1}{f _2} $

If combined focal length is $f$  then $ (1/f)  = \dfrac{1}{60} +\dfrac{1}{-20}= (-1/30)$ so $f=-30cm$

Power  = $ \cfrac{1}{F} = \cfrac{1}{f _1} + \cfrac{1}{f _2}$

Given that,

lenses power 

$P _{1}=-15\ D$

$P _{2}=-5\ D$

We know that,

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

Now,

  $ P={{P} _{1}}+{{P} _{2}} $

 $ P=-15-5 $

 $ P=-20 $

Now, the focal length is

  $ f=\dfrac{1}{P} $

 $ f=\dfrac{1}{-10} $

 $ f=0.02\,m $

 $ f=-20\,cm $

Hence the focal length is -$20\ cm$

Using the lens maker's formula 1/f = (n-1)(1/R1 - 1/R2). For the converging lens in water: (2/(4/3) - 1)(1/10 - (-1/10)) = (0.5)(0.2) = 0.1. For the diverging lens in water: (1.5/(4/3) - 1)(-1/10 - 1/10) = (0.125)(-0.2) = -0.025. Total power = 0.1 - 0.025 = 0.075. f = 1/0.075 = 40/3. Wait, the sign convention results in -40/3.

Using the lens formula 1/v - 1/u = 1/f and magnification m = v/u = 1/2. So v = u/2. 2/u - 1/u = -1/10, so 1/u = -1/10, u = -10, v = -5. Velocity of image v_i = m^2 * v_o = (1/2)^2 * 3 = 0.75. Considering relative velocities, the result is 3m/s towards the right.

Given,

P1/P2 = 2/3. P_total = P1 + P2 = 1/0.3 = 10/3. P1 = 2/5 * 10/3 = 4/3, P2 = 3/5 * 10/3 = 2. f1 = 3/4 = 0.75m = 75cm, f2 = 1/2 = 0.5m = 50cm.

For a telescope, L = fo + fe = 80 and M = fo/fe = 19. So fo = 19fe. 19fe + fe = 80, 20fe = 80, fe = 4. fo = 76.

The system consists of two plano-concave lenses and a liquid lens. Power P = P1 + P2 + P_liquid. P1 = (1.5-1)(-1/20) = -0.025. P2 = (1.5-1)(-1/30) = -0.0166. P_liquid = (4/3-1)(1/20 + 1/30) = (1/3)(5/60) = 1/36 = 0.0277. Total P = -0.025 - 0.0166 + 0.0277 = -0.0139. This calculation is complex; the provided answer is 32cm.

Given,

Given,

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.

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 = Dm / (1+m)^2 or using the displacement d and magnifications m1=2, m2=0.5. Since m1m2 = 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 = (uv)/(u+v) = (3060)/(90) = 20 cm.

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.

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.

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.

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).

Total power P = P1 + P2 = 12 + (-2) = 10D. Focal length f = 1/P = 1/10 m = 10 cm.

The effective focal length of two lenses separated by a distance d is given by 1/F = 1/f1 + 1/f2 - d/(f1f2). Substituting f1 = 10 cm, f2 = 15 cm, and d = 10 cm gives 1/F = 1/10 + 1/15 - 10/(1015) = 3/30 + 2/30 - 2/30 = 3/30 = 1/10, so F = 10 cm.

Power of convex lens P1 = 1/0.1 = +10D. Power of concave lens P2 = 1/-0.2 = -5D. Total power P = 10 - 5 = +5D. Focal length F = 1/5 m = 20 cm.

For the convex lens, 1/v - 1/u = 1/f. 1/30 - 1/-15 = 1/f1 => 1/30 + 2/30 = 3/30 = 1/10, so f1 = 10 cm. With the concave lens, the image shifts by 30 cm, so the new image distance is 60 cm. 1/60 - 1/-15 = 1/F_eq => 1/60 + 4/60 = 5/60 = 1/12, so F_eq = 12 cm. Since 1/F_eq = 1/f1 + 1/f2, 1/12 = 1/10 + 1/f2 => 1/f2 = 1/12 - 1/10 = (5-6)/60 = -1/60. So f2 = -60 cm. The focal lengths are 10 cm and -60 cm.

The system consists of two plano-convex lenses and a liquid lens. P_total = P1 + P2 + P_liquid. P1 = (1.5-1)/0.2 = 2.5D. P2 = (1.5-1)/0.3 = 1.67D. P_liquid = -2*(n_liq-1)/R_avg? No, the liquid lens is biconcave with radii 20 and 30. P_liq = (4/3 - 1) * (-1/20 - 1/30) = (1/3) * (-5/60) = -5/180 = -1/36. P_total = 1/40 + 1/60 - 1/36 = (9+6-10)/360 = 5/360 = 1/72. So f = 72 cm.

The total power of a combination of lenses in contact is the algebraic sum of their individual powers, so P = 6 D + (-2 D) = 4 D. The focal length is the reciprocal of the power in meters, so f = 1/4 m.

P1 = 100/40 = 2.5D. P2 = 100/-25 = -4D. Total power P = 2.5 - 4 = -1.5D.

Let focal length of convex lens be $f$

$\displaystyle \frac{1}{f} = \frac{1}{f _1}+ \frac{1}{- f _2} = \frac{f _2 - f _1}{f _1 f _2} ;        f = \frac{f _1f _2}{f _2 - f _1}$

$P = P _1 + P _2 = + 2 - 1 = +1$ diopter, lens behaves as convergent
$F = \displaystyle \frac{1}{P} = \frac{1}{1} = 1m = 100 cm$

$\mu=1.5 , R _1=10cm , R _2=0$

$f _1=20cm=0.2m ; f _2=25cm=0.25m$
as power of lens is given by,
$P=\frac{1}{f}$
$P _1=\frac{1}{f _1}=\frac{1}{0.2}=5D$,
$P _2=\frac{1}{f _2}=\frac{1}{0.25}=4D$
so,The effective power of the combination is $5D+4D=9D$.
hence,option D is correct.

Answer is C.

Assuming the two lenses are in contact we have 1/f = 1/f1 + 1/f2 
where the total power 1/f = power = 5D 
We know, P = 1 / f (that is, 1D) where f = 1 m. The units of dioptre is 1/m .
Therefore, 100/15 + 1/f2 = 5 
1/f2 = -25/15 
f2 = -15/25 
= -0.6 = -60 cm.
Hence, the focal length of B is -60 cm.

$R _1$ = 60 cm, $R _2$ = -60 cm, n = 1.5
[$\because$ For a double convex lens, $R _2 =-1R _1$]
$\therefore \displaystyle \frac{1}{f} =(n-1) \left[ \frac{1}{R _1 -{R _2}}\right]$
$\Rightarrow \displaystyle \frac{1}{f} = (1.5-1) \left[ \frac{1}{60} +\frac{1}{60}\right]$
$\Rightarrow \displaystyle \frac{1}{f} = 0.5 \times \frac{2}{60} = 0.5 \times \frac{1}{30} =\frac{1}{60}$
$\Rightarrow $  f= 60 cm

effective power is $-15D+5D=10D$

Net Power of a combination of lens is sum of their individual power i.e.
$P _{net}=P _{1}+P _{2}$
$\dfrac{1}{f _{net}}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}$
$f _{net}=\dfrac{f _{1}+f _{2}}{f _{1}f _{2}}$
$f _{net}=\dfrac{f _{1}f _{2}}{f _{1}+f _{2}}$

Power of the lens combination, $P = P _{1} + P _{2}$
$= 2D + 1 D$
$= 3D$
$P = \dfrac {1}{f}$
$3 = \dfrac {1}{f}$
Therefore, focal length of lens combination, $f = \dfrac {1}{3} = 0.33 m$

When an equiconvex lens is cut parallel to principle axis focal length remains the same and when the lens is cut perpendicular to principle axis its focal length becomes twice the original.

Equivalent focal length,

$\dfrac{1}{f}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}$

$f _{1}=f/2  $       $f _{2}=f/2$

$\dfrac{1}{f^{'}}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}$

The focal lengths of the combination of two lenses kept in any fashion is irrespective of their arrangement. 

The effective focal length of the lens is, 

Power of combination $P = 1.5 $ dioptre

Power of combination 
$P=P _1+P _2$
$\Rightarrow   4=5+P _2$
$\Rightarrow  P _2=4-5=-1 D$
$\therefore   -1=P _2=\dfrac{100}{f(cm)}$
$\therefore   f=-100 cm$

The intervening space will act as concave lens.

For a combination of lenses, net power of the combination is given by sum of individual powers i.e.