Fill in the blanks for convex lens.
f(m) 0.2 . . . 0.1
P (D) . . . 2 . . .
$h_{1}$ . . . 5 10
$h_{2}$ -30 -20 . . .
M -2 . . . -0.5

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Solution

we know $P o w e r (P) = \frac{1}{f o c a l l e n g t h (f)}$
and magnification(M) $= \frac{h e i g h t o f i m a g e (h_{2})}{h e i g h t o f i m a g e (h_{1})}$
so,the solved table is:
f(m) :0.2 0.5 0.1
P(D):5 2 10
$h_{1}$ :15 5 10
$h_{1}$ :-30 -20 5
M:-2 -4 -0.5

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Similar questions

Fill in the blanks for convex lens.

f (m)	0.2	…	0.1
P (D)	…	2	…
h₁	…	5	10
h₂	−30	−20	…
M	−2	…	−0.5

What will be the area A of the quadrilateral in each case with their diagonal $d c m$ & the sum of their heights $(h_{1} + h_{2}) c m$ .

i) $d = 20, (h_{1} + h_{2}) = 10$

ii) $d = 15, (h_{1} + h_{2}) = 20$

iii) $d = 20, (h_{1} + h_{2}) = 25$

iv) $d = 40, (h_{1} + h_{2}) = 30$

Two block of wood of masses $m_{1}$ and $m_{2}$ and each of specific gravity 0.5 are submerged at a depth of $h_{1}$ and $h_{2}$ in
a vessel $(h_{2} > h_{1})$ filled with water, which is accelerated upwards with an acceleration $\frac{g}{2}$ . the difference in
time taken by the blocks to reach the surface, when released with zero velocity is :

Q. If

H_{1}, H_{2}, \dots, H_{.20}

20

harmonic means between

2

and

3,

then

\frac{H_{1} + 2}{H_{1} - 2} + \frac{H_{20} + 3}{H_{20} - 3} = ?

Q. If

H_{1}, H_{2}, \dots \dots, H_{20}

be 20 harmonic means between

2

and

3

, then the value of

\frac{H_{1} + 2}{H_{1} - 2} + \frac{H_{20} + 3}{H_{20} - 3}

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Fill in the blanks for convex lens.f(m)0.2. . .0.1P (D). . .2. . .h1. . . 510h2-30-20. . .M-2. . .-0.5

we know Power(P)=1focallength(f)and magnification(M)=height of image(h2)height of image(h1)so,the solved table is:f(m) :0.2 0.5 0.1P(D):5 2 10h1:15 5 10h1:-30 -20 5 M:-2 -4 -0.5

Fill in the blanks for convex lens.
f(m) 0.2 . . . 0.1
P (D) . . . 2 . . .
$h_{1}$ . . . 5 10
$h_{2}$ -30 -20 . . .
M -2 . . . -0.5

we know $P o w e r (P) = \frac{1}{f o c a l l e n g t h (f)}$
and magnification(M) $= \frac{h e i g h t o f i m a g e (h_{2})}{h e i g h t o f i m a g e (h_{1})}$
so,the solved table is:
f(m) :0.2 0.5 0.1
P(D):5 2 10
$h_{1}$ :15 5 10
$h_{1}$ :-30 -20 5
M:-2 -4 -0.5